College textbook business math and statistics - section b - business mathematics - ch 7 - calculus-i

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Chapter 7: Calculus-I
7.0 Introduction:
In real life business/ monetary situations, calculus is used to find the rate of change of quantities,
or to know their maxima and minima. We could even use calculus in finding the rate of change
of velocity and acceleration of objects (application of this concept in Physics). Calculus is used
specially to find the slope of two dimensional curves at given points, the length, and volume and
surface areas of different three dimensional solids.
With calculus, we have the ability to find the effects of changing conditions on a system. By
studying these, we can learn how to control a system to make it do what one wants it to do.
Because of the ability to model and control systems, calculus gives us extraordinary power over
the material world. Calculus is the language of engineers, scientists and economists. This has a
huge impact on our daily lives – from use in microwaves, cell phones, television, and car to
applications in the fields of medicine, business, Finance, economics, and even national defense.
Calculus is used by various people such as biologists, engineers, architect, space flight engineers,
statisticians, physicists etc.
7.1 Objective:
The purpose of calculus is to learn Limits and Continuity, Differentiation, Application of
Derivatives, Integration, Integrals and Transcendental Functions.
In this chapter we will study about the concept and rules of Differentiation, Rate of change of a
function, Increasing and decreasing functions, Application of derivatives, criteria for finding the
maximum and minimum of a function, about concavity and convexity (on the graph) and its
conditions, point of inflection and conditions for it to exist for a function.
7.2.0 Concept and rule of differentiation
The procedure that we use to find the derivative of a function is called Differentiation.
Differentiation is the process of studying change (for a mathematical function). For example, the
acceleration of a moving object can be found by differentiating its velocity. It is one of the most
important topics in calculus.
7.2.1 Introduction
The derivative of the function  y f x , can be written as  
dy
f x
dx
 . It represents the rate of
change of the function with respect to x . The function  f x is said to be differentiable at x c , if
the limit
   
0
lim
h
f c h f c
h
 
exits. This limit is called derivative of the function  f x at x c .

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The limit can be denoted as  f c or
x c
dy
dx 
 
 
 
. The function  y f x is said to be differentiable
in the interval  ,a b if it is differentiable at each and every point in this interval. A differentiable
function is always continuous at that point.
Example 1: Find the derivative of the function   2
16f x x by definition.
Solution: The difference quotient for  f x is
     
2 2
16 16f x h f x x h x
h h
   

 2 2 2
2 2 2
2
16 2 16
16 32 16 16
32 16
32 16
x xh h x
h
x xh h x
h
xh h
h
x h
  

  



 
Therefore, the derivative of the above function is
 
   
 
0
0
lim
lim 32 16
32
h
h
f x h f x
f x
h
x h
x


 
 
 

Example 2: Find the derivative of the function   3
f x x and the slope of the tangent line to the
curve 3
y x at the point where 1x   . Then find the equation of the tangent line at this point.
Solution: The difference quotient for  f x is
     
3 3
f x h f x x h x
h h
   

 3 2 2 3 3
3 2 2 3 3
2 2 3
2 2
3 3
3 3
3 3
3 3
x x h xh h x
h
x x h xh h x
h
x h xh h
h
x xh h
   

   

 

  
Therefore, the derivative of the above function is

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 
   
 
0
2 2
0
2
lim
lim 3 3
3
h
h
f x h f x
f x
h
x xh h
x


 
 
  

Therefore, the slope of the tangent line at the point 1x   is  1f   and its value is
   
2
1 3 1
3
f    

To find the equation of the tangent line we also need the y - coordinate. This can be obtained by
substituting the value 1x   in the function   3
f x x as
   
3
1 1
1
f   
 
The point is 1, 1  . Now we get the equation of the tangent line in slope point form as
    1 3 1
1 3 3
3 2
y x
y x
y x
    
  
 
Therefore, the equation of the tangent line is 3 2y x 
Example 3: Find the slope of the tangent to the curve  y f x x  at the point 4,2 .
Solution: The slope of the tangent to the curve  y f x x  at the point 4,2 is  4f 
We get  4f  as
 
   
0
0
0
0
4 4
4 lim
4 4
lim
4 2
lim
4 2 4 2
lim .
4 2
h
h
h
h
f h f
f
h
h
h
h
h
h h
h h




 
 
 

 

    
     
0
0
0
4 4
lim
4 2
lim
4 2
1
lim
4 2
h
h
h
h
h h
h
h h
h



  
  
  
 
  
  
 
  
  

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1
4 2
1
4



Hence, slope of the tangent to the curve  y f x x  at the point 4,2 is
1
4
.
Following are the rules for differentiation
Rule I: If c is a constant then   0
d
c
dx

Example 4: Find the derivative of 2y  .
Solution: The derivative of 2y  is
 5
0
dy d
dx dx


Rule 2: If n is any real number then   1n nd
x nx
dx


Example 5: Differentiate each of the following
(i)
5
2
x (ii) 3
x (iii)
2
x
(iv) 2
1
x
Solution: By using Rule 2 we get
(i) The derivative of
5
2
x is
5 5
1
2 2
5
2
d
x x
dx
 
 
 
3
2
5
2
x
(ii) The derivative of 3
x is
 3 2 1
3
3
d
x x
dx
x



(iii) The derivative of
2
x
is

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1
2
1
1
2
3
2
2
2
1
2
2
d d
x
dx dxx
x
x
 
 
 
 

  
        
 
  
 
 
(iv) The derivative of 2
1
x
is
 2
2
2 1
3
1
2
2
d d
x
dx x dx
x
x

 

 
 
 
 
 
Rule 3: If f is a differentiable function and c is a constant then
   .
d d
c f x c f x
dx dx
      
Example 6: Differentiate the following
(i) 3
6x (ii)
5
3
7
3
x (iii) 5
500
x
Solution: By using the Rule 3
(i) The derivative of 3
6x is
 3 3
2
6 6
18
d d
x x
dx dx
x


(ii) The derivative of
5
3
7
3
x is
5 5
1
3 3
2
3
7 7 5
3 3 3
35
9
d d
x x
dx dx
x
   
   
   

(iii) The derivative of 5
500
x
is
1
5
5
500
500
d d
x
dx dxx
  
   
   
1
1
5
1
6
1
500
5
100
x
x
 

 
  
 
 

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Rule 4: If f , g are differentiable functions and c is a constant then
       
d d d
f x g x f x g x
dx dx dx
            
Example 7: Differentiate the following
(i) 3
x x (ii)
2
1x
x

(iii) 4 3 2
3 8 3x x x x   
Solution: By using the Rule 4
(i) The derivative of 3
x x is
 3 3
2
3 1
d d d
x x x x
dx dx dx
x
  
 
(ii) The derivative of
2
1x
x

is
2
1
1 1
1 1
1
d x d
x
dx x dx x
d d
x x
dx dx
x

 
   
    
  
 
  
2
1
1
x
 
(iii) The derivative of 4 3 2
3 8 3x x x x    is
           4 3 2 4 3 2
3 2
3 8 3 3 8 3
4 9 16 1
d d d d d d
x x x x x x x x
dx dx dx dx dx dx
x x x
        
   
Rule 5: Product Rule
If f , g are differentiable functions then
           
d d d
f x g x f x g x g x f x
dx dx dx
           
Example 8: If   3 4 3
4 7 2 1y x x x x     , find
dy
dx
Solution: By using the Product rule we get
  
       
3 4 3
3 4 3 4 3 3
4 7 2 1
4 7 2 1 2 1 4 7
dy d
x x x x
dx dx
d d
x x x x x x x x
dx dx
      
         

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     3 3 2 4 3 2
6 5 4 3 3 2 6 4 5 3 2
6 5 4 3 2
4 7 4 8 2 1 3 4
4 8 16 32 28 56 3 4 6 8 3 4
7 14 20 40 53 4
x x x x x x x
x x x x x x x x x x x
x x x x x
       
           
     
Rule 6: Quotient Rule
If f , g are differentiable functions then
 
 
       
 
2
d d
g x f x f x g xf xd dx dx
dx g x g x
       
 
    
Example 9: If
2 1
1
x
y
x



, find
dy
dx
Solution: By using the Quotient rule we get
       
 
2
3 1
4
1 3 1 3 1 4
4
dy d x
dx dx x
d d
x x x x
dx dx
x
 
   
    


     
 
 
 
2
2
2
1 3 3 1 1
4
3 2 3 1
4
3
4
x x
x
x x
x
x
  


  


 

Example 10: If
3 2
2
2 3 1
1
x x
y
x
 


, find
dy
dx
Solution: By using the Quotient rule we get

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       
 
     
 
 
3 2
2
2 3 2 3 2 2
22
2 2 3 2
22
4 3 2 4 3
22
2 3 1
1
1 2 3 1 2 3 1 1
1
1 6 6 2 3 1 2
1
6 6 6 6 4 6 2
1
dy d x x
dx dx x
d d
x x x x x x
dx dx
x
x x x x x x
x
x x x x x x x
x
  
   
      


    


     


 
4
22
2 4
1
x x
x



Rule 7: Chain Rule
If  y f u is a differentiable function in u and  u g x is a differentiable function in x , then
 y f g x    is a differentiable function in x and
dy dy du
dx du dx

Example 11: If 2
y u and 3
5 1u x x   , find
dy
dx
Solution: By using Chain rule,
   
 
  
2 3
2
3 2
5 1
2 3 5
2 5 1 3 5
dy dy du
dx du dx
d d
u x x
du dx
u x
x x x

  
 
   
Example 12: If 3
y u and 3
4 8u x x   , find
dy
dx
Solution: By using Chain rule

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 
 
 
1
33
1
1
23
2
23
5 1
1
3 5
3
1
3 5
3
dy dy du
dx du dx
d d
u x x
du dx
u x
u x



 
   
 
 
 
   
2
3 23
1
4 8 3 5
3
x x x

   
Example 13: If  
252
2 7 1y x x   , find
dy
dx
Solution: The function y can be written as 25
y u , where 2
2 7 1u x x  
Thus by Chain rule,
dy dy du
dx du dx

   
 
   
25 2
24
242
2 7 1
25 2 7
25 2 7 1 2 7
d d
u x x
du dx
u x
x x x
  
 
   
Example 14: If
4
2 1
7
x
y
x
 
  
 
, find
dy
dx
Solution: The function y can be written as 4
y u , where
2 1
7
x
u
x



Thus by Chain rule and Quotient rule
 
       
 
   
 
4
3
2
3
2
2 1
7
7 2 1 2 1 7
4
7
7 2 2 1 12 1
4
7 7
dy dy du
dx du dx
d d x
u
du dx x
d d
x x x x
dx dxu
x
x xx
x x

 
  
 
 
     
  
 
 
    
   
    

10. Evelyn
 
 
 
 
3
2
3
2
3
5
2 1 2 14 2 1
4
7 7
2 1 15
4
7 7
2 1
60
7
x x x
x x
x
x x
x
x
     
   
    
   
        

 

TIPS AND TRICKS #1
Let n
y u and  u f x . Applying the Chain rule
    
 
   
-1
1
.
n
n
n
dy dy du
dx du dx
d d
u f x
du dx
nu f x
n f x f x

 
 
 
   
Thus,
     
1n nd
f x n f x f x
dx

       
Example 15: If  
95 2
7y x x   , find
dy
dx
Solution: By applying above rule we can solve as
 
   
   
95 2
85 2 5 2
85 2 4
7
9 7 7
9 7 5 2
dy d
x x
dx dx
d
x x x x
dx
x x x x
  
    
   
Derivative of logarithmic and exponential functions:
(1) Derivative of loga x
 
1
log loga a
d
x x
dx x

On replacing a by e above to get
 e
1
log
d
x
dx x

Example 16: If 35logy x , find
dy
dx

11. Evelyn
Solution: By applying above rule we can solve as
35 log
35
dy d
x
dx dx
x


TIPS AND TRICKS #2
If logy u then to get the derivative we use the chain rule as
dy dy du
dx du dx

 log
1
dy d du
u
dx du dx
du
u dx


Example 17: If  2
log 5 6y x  , find
dy
dx
Solution: By applying above rule we can solve
Consider 2
5 6u x 
    
 
 
2 2
2
2
log 5 6 5 6
1
10
5 6
10
5 6
dy d d
x x
dx du dx
x
x
x
x
  




Example 18: If  2
log 2 7y x x  , find
dy
dx
Solution: We use the product rule to solve, we get
   
   
 
2 2
2
2
log 2 7 log 2 7
2 7 2 log 2 7
2 7
2
2 log 2 7
2 7
dy d d
x x x x
dx dx dx
x d
x x x
x dx
x
x x
x
   
   

  

Example 19: If
log x
y
x
 , find
dy
dx
Solution: We use the quotient rule to find the derivative of
log x
y
x


12. Evelyn
2
2
2
log log
1
log
1 log
d d
x x x x
dy dx dx
dx x
x x
x
x
x
x


 
 
 


Example 20: If  2
log 1y x x   , find
dy
dx
Solution: Consider 2
1u x x   then we can find
dy
dx
as
 
 
2
2
1
2 2
2
log
1
1
1
1
1
1
1
dy d
u
dx dx
du
u dx
d
x x
dxx x
d d
x x
dx dxx x


  
 
 
   
   
   
 
1
12 22
2
1
2 2
2
1 1
1 1 1
21
1 1
1 1 2
21
d
x x
dxx x
x x
x x


  
     
    
  
    
    
2 2
2
2 2
2
1
1
1 1
1 1
1 1
1
1
x
x x x
x x
x x x
x
 
  
   
  
  
    


(2) Derivative of , 0x
a a 
  logex x ad
a a
dx

On replacing a by e above we get

13. Evelyn
 e ex xd
dx

TIPS AND TRICKS #3
If eu
y  then we get the derivative by using the chain rule as
dy dy du
dx du dx

 e
e
u
u
dy d du
dx du dx
du
dx


Example 21: If
2
2 5
ex x
y  
 , find
dy
dx
Solution: By applying above rule we can solve,
Consider 2
2 5u x x  
dy dy du
dx du dx

   
 
2
2
2 5 2
2 5
e 2 5
e 2 2
x x
x x
dy d
x x
dx dx
x
 
 
  
 
Example 22: If 2
2 2logx
y x x   , find
dy
dx
Solution: We can write
dy
dx
as
2
2 2 log
2
2 2 log 2
x
x
dy d d d
x x
dx dx dx dx
x
x
  
  
Example 23: If
2
5 7 1
e x x
y  
 , find
dy
dx
Solution: By applying above rule we can solve,
Consider 2
5 7 1u x x  
dy dy du
dx du dx

   
 
2
2
5 7 1 2
5 7 1
e 5 7 1
e 10 7
x x
x x
dy d
x x
dx dx
x
 
 
  
 

14. Evelyn
7.2.2 Rate of change of a function
The rate of change of a function is calculated by differentiating the function. For example
dy
dx
denotes the rate of change in y with respect to x . For a linear function such
as  f x mx c  , changes at a constant rate m with respect to the independent variable x . The
rate of change of  f x is constant and is called its slope. For a non-linear function  f x , the rate
of change is not constant and varies with respect to x .
If the position of an object moving at time t along a straight line is given by  s t , then the object
has velocity is represented as rate of change of  s t and it is given as    
ds
v t s t
dt
  and
acceleration is represented as rate of change of velocity  v t and it is given
as      
2
2
dv d s
a t v t s t
dt dt
     .
Example 24:   2
3 5 2s t t t   describes a particle motion with units in meters find
(i) Velocity and acceleration functions and
(ii) The rate of change of velocity and acceleration at 4 sect 
Solution:
(i) The derivative of  s t gives the velocity  v t .
Hence,
   
 2
3 5 2
6 5
d
v t s t
dt
d
t t
dt
t

  
 
Now, the derivative of velocity  v t gives the acceleration
   
 6 5
6
d
a t v t
dt
d
t
dt

 

(ii) We get the instantaneous velocity and acceleration at 4 sect  by letting 4 sect  in the
above equations
   
 
4 6 4 5
24 5
29
4 6
v
a
 
 



15. Evelyn
Example 25: Find the rate of change of   2
2 12 5f x x x   with respect to x . Evaluate rate of
change when 3x 
Solution:
The rate of change of  f x with respect to x is:
   2
2 12 5
4 12
d d
f x x x
dx dx
x
  
 
When 3x 
 
3
4 3 12
0
x
df
dx 
 

Example 26: Show that the rate of change of the area of a circle with respect to its radius is
equal to circumference.
Solution: Let A be the area of the circle with radius r . Then
2
A r
Thus the rate of change of area with respect to radius is
2
2
dA d
r
dr dr
r




Therefore the above value 2 r represents the circumference of a circle.
Exercise 7.2:
1. Find the derivative of the functions by using the definition
(i)   2f x x (ii)   5 4f x x 
(iii)   2
5 12f x x x   (iv)   2
9f x x 
2. Find the slope of the tangent to the curve   2
4y f x x   at 1x  .
3. Find the slope of the tangent to the curve   2
2 10y f x x x    at 2x  .
4. Find the derivative of the following functions
(i)   5
3f x x (ii)   7
f x x (iii)   7 4f x x 
(iv)  
5 8
16
x
f x

 (v)   2
ef x 
5. Differentiate the following functions
(i)   3 2
3 7 7f x x x   (ii)  
4
2 3
2 2
2 4
x
f x x x    
(iii)  
3
7
2
x x
f x
x


6. Differentiate with respect to x the following functions

16. Evelyn
(i) 6 4 2
3 2x x x
  (ii)   m m
f x x a  (iii)  
2
1
f x x
x
 
  
 
7. Find the derivatives of the following functions
(i)   1 1x x  (ii)   2 2
3 7 2 5 9x x x x    (iii) 2
logx x
(iv) 3 2
e x
x (v) 3
3x
x (vi) 5 2
e logx
x x
(vii)  2
16 ex
x x (viii) log 2x
x x 
8. Differentiate with respect to x each of the following functions
(i)
1
1
x
x


(ii)
2
2
1
1
x x
x x
 
 
(iii)
5
1 3x
(iv) 2
log x
x
(v)
2
1
x
x
(vi)
2
1
log
1
x
x
 
 
 
(vii)
2
e
log
x
x
(viii)
3
2
2 1
3 1
x
x


9. Find the derivatives of the following functions
(i)
1
1 1x x  
(ii)
2 2 2 2
1
x a x b  
(iii)
a x a x
a x a x
  
  
(iv)
2 2
2 2
1 1
1 1
x x
x x
  
  
10. Use chain rule to find
dy
dx
of each of the following functions
(i) 2
2 7y u u  and 3
5u x x 
(ii) 2
y u and
1
1
x
u
x



(iii) 2
3 8 4y u u   and logu x
(iv) eu
y  and 6 2
1u x x  
(v) logy u and
1
u x
x
 
11. Find the derivatives of the following functions
(i)  
5
5 4x  (ii)  
503 2
2 8x x x  (iii) 2
7
2 1x x 
(iv)  log log x (v)  log log log x   (vi) 3
log x
(vii)  
3
log x (viii)  2
log 3 2 5x x  (ix)  log ex
x
(x)  
32
log 4 5x x 
12. Differentiate with respect to x each of the following functions
(i) 3 1
4 x
(ii)  
43
e 1x
 (iii) 1+
e x
(iv)
1
ex
(v)
3
2 5
4 x x
(vi)
2
e logx
x

17. Evelyn
13. (i) If 2
log 1y x x   
 
, prove that
2
1
1
dy
dx x
 

(ii) If 1 1y x x    , prove that 2 1
1
2
dy
x y
dx
 
(iii) If 2
1
m
y x x   
 
, prove that
2
1
dy my
dx x


(iv) If
2
x
y
x


, prove that  1
dy
x y y
dx
 
(v) If
1
y x
x
  , prove that
1
2
dy
x x
dx x
 
Answers to Exercise 7.2:
1. (i) 2 (ii) 5
(iii) 2 5x  (iv) 2x
2. 2
3. 7
4. (i) 4
15x (ii) 6
7x
(iii) 7 (iv)
5
16
(v) 0
5. (i) 2
9 14x x (ii) 33
4
2
x x  
(iii)
35 1
4 4
x x
x

6. (i) 5 3 3
18 4 4x x x
  (ii) 1m
mx 
(iii) 3
1
2 x
x
 
 
 
7. (i) 2 2x  (ii) 3 2
8 3 16 8x x x  
(iii)  1 2logx x (iv)   2 2
2 3 e x
x x
(v)   2
3 log3 3x
x x (vi)  4 2
e 1 2 log 5logx
x x x x 
(vii)  2
16 31 1 ex
x x  (viii) 1 log 2 log2x
x 
8. (i)
 
2
2
1x 
(ii)
 
2
22
2 2
1
x
x x

 
(iii)
 
2
15
1 3x
(iv) 3
1 2log x
x

(v)
 
3
2 2
1
1 x
(vi)
  
2
2
1 2
1 1
x x
x x
 
 
(vii)
 
 
2
2
e 2 log 1
log
x
x x
x x

(viii)
 
 
3
22
6 1
3 1
x x x
x
 


18. Evelyn
9. (i)
1 1 1
4 1 1x x
 
 
  
(ii)
 2 2 2 2 2 2
1 1x
a b x a x b
 
 
   
(iii)
2 2
2 2 2
a a x a
x a x
   
 

(iv)
4
3 4
2 1 1
1
x
x x
  
 

10. (i)   3 2
20 4 7 5 3x x x  
(ii)
 
 
3
4 1
1
x
x
 

(iii)
6log 8x
x

(iv)  
6 2
5 1
6 2 ex x
x x  

(v)
 
 
1
2 1
x
x x


11. (i)  
4
25 5 4x  (ii)   
492 3 2
100 3 16 1 8x x x x x   
(iii)
 
 
22
7 4 1
2 1
x
x x
 
 
(iv)
1
logx x
(v)
 
1
log log logx x x
(vi)
3
x
(vii)
 
2
3 log x
x
(viii) 2
6 2
3 2 5
x
x x

 
(ix)
1x
x

(x) 2
6 12
4 5
x
x x

 
12. (i)   3 1
3log4 4 x
(ii)  
33 3 1
12e ex x
(iii)
1+
e
2
x
x
(iv)
1
2
ex
x

(v)  
3
2 2 5
log4 6 5 4 x x
x x 
 (vi)
2 1
e 2 logx
x x
x
 
  

19. Evelyn
7.3.0 Maxima and Minima involving second or Higher derivatives
Consider a differentiable function in x as  y f x . Now,  
dy
f x
dx
 is called the first
derivative of f . Differentiating the first derivative again with respect to x , we get the second
derivative of f and it is denoted as  f x . Similarly, on differentiating the second derivative
again if it is differentiable, we get the third derivative and it is written as  f x .
The above explanation can be represented as follows:
 
 
 
1
2 2
2 2 2
3 2
3 3 2
First Derivative : , , ,
Second Derivative : , , ,
Third Derivative : , , ,
dy d
y y f x
dx dx
d y d
y y f x
dx dx
d y d
y y f x
dx dx
   
   
   
7.3.1 Increasing and Decreasing functions
Consider a function  y f x in the interval  ,a b the function f is said to be increasing if
 f x increases as x increases on the interval  ,a b and it is said to be decreasing if
 f x decreases as x decreases on the interval ,a b .
Therefore we can say that f is increasing on  ,a b if    1 2f x f x whenever 1 2x x with
1x and 2x in  ,a b and f is decreasing on  ,a b if    1 2f x f x whenever 1 2x x with 1x and 2x
in ,a b .
7.3.2 Sign of derivative
Let us consider a differentiable function in the interval  ,a b as  y f x and let x be a point in
the same interval. Now,  f x is considered to be positive or increasing if the derivative of it
greater than zero, that is   0f x  for all values of x in the interval  ,a b and  f x is
considered to be negative or decreasing if the derivative of it less than zero, that is   0f x  for
all values of x in the interval  ,a b
Example 27: Find the intervals on which the function   3 2
2 9 12f x x x x   is increasing or
decreasing.
Solution: Differentiating the function   3 2
2 9 12f x x x x   gives

20. Evelyn
 
  
2
6 18 12
6 1 2
f x x x
x x
   
  
To determine when  f x is positive and when  f x is negative, we find the sign the of
 f x by considering the intervals determined by the roots of   6 1 2x x  namely 1 and2 .
If 1x  , then     6 0f x     so f is increasing;
If1 2x  , then     6 0f x     and so f is decreasing;
If 2x  , then     6 0f x     and so f is increasing;
Therefore we can say that f is increasing on  ,1 and 2, , and is decreasing on 1,2 .
Example 28: Find the intervals on which the function   3 2
2 3 12f x x x x    is increasing or
decreasing.
Solution: Differentiating the function   3 2
2 3 12f x x x x    gives
 
  
2
6 6 12
6 1 2
f x x x
x x
    
   
To determine when  f x is positive and when  f x is negative, we find the sign of  f x by
considering the intervals determined by the roots of   6 1 2x x   namely 1 and2 .
If 1x   , then         0f x        so f is decreasing on , 1  ;
If 1 2x   , then         0f x        and so f is increasing on 1,2 ;
If 2x  , then     6 0f x     and so f is decreasing on 2, ;
Therefore we can say that f is increasing on 1,2 , and f is decreasing on the intervals
 , 1  and 2, .
Example 29: Find the intervals on which the function   4 3 2
8 22 24 6f x x x x x     is
increasing or decreasing. [C.A Foundation May 1994]
Solution: Differentiating the function   4 3 2
8 22 24 6f x x x x x     gives
 
   
3 2
4 24 44 24
4 1 2 3
f x x x x
x x x
    
   

21. Evelyn
To determine when  f x is positive and when  f x is negative, we find the sign of  f x by
considering the intervals determined by the roots of    4 1 2 3x x x   , namely1, 2 and3 .
If 1x  , then        4 0f x        so f is decreasing on ,1 ;
If1 2x  , then        4 0f x        and so f is increasing on 1,2 ;
If2 3x  , then        4 0f x        and so f is decreasing on 2,3 ;
If 3x  , then        4 0f x        and so f is increasing on 3, ;
Therefore we can say that f is increasing on  1,2 and 3, , and f is decreasing on the
intervals ,1 and 2,3 .
7.3.3 Stationary value
Consider a differentiable function  y f x in the interval ,a b , let 0x be a point which lies in
the same interval then the point 0x in the interval of f at which  0 0f x  is called a stationary
point of f . We get the stationary value by substituting the stationary point in the
function  y f x , that is, the value  0y f x is the stationary value.
7.3.4 Criteria for maxima and minima
To understand the criteria for maxima and minima, we study increasing and decreasing
functions. A differentiable function  f x is said to be increasing if its derivative is greater than
zero, that is,   0f x  for all values of x in the interval  ,a b ; and it is said to be decreasing if
its derivative is less than zero, that is,   0f x  for all values of x in the interval  ,a b .
Now, consider a differentiable function  f x in the interval  ,a b and  ,c a b then
(a) The function  f x has a maximum value at x c , if
(i)   0f c  and
(ii)   0f c  provided  f c exists
(b) The function  f x has a minimum value at x c , if
(i)   0f c  and
(ii)   0f c  provided  f c exists
We should also understand the concept of local maximum and local minimum.

22. Evelyn
Local maximum:
Let  y f x be a real-valued function that is defined in some interval containing a point 0x x .
Then the function  f x is said to have a local maximum at 0x x if    0f x f x for all values
of x sufficiently close to 0x . The local maximum value is  0f x .
Local minimum:
Let  y f x be a real-valued function that is defined in some interval containing a point 0x x .
Then the function  f x is said to have a local minimum at 0x x if    0f x f x for all values
of x sufficiently close to 0x . The local minimum value is  0f x .
We should also know that the function  f x has a local maximum at 0x x if  f x changes
sign from positive to negative as x increases through 0x and has a local minimum at 0x x if
 f x changes sign from negative to positive as x increases through 0x .
Stationary Point: A point 0x in the domain of f at which  0 0f x  is called a stationary point
of f .
Example 30: Use the second derivative test to find the local maxima and minima of
  3 2
2 15 36 18f x x x x   
Solution: Differentiating the given function with respect to x , we obtain
 
  
2
6 30 36
6 2 3
f x x x
x x
   
  
Now, equate   0f x 
Hence,
  6 2 3 0x x  
This implies that
2 and 3x x 
Also
  12 30f x x  
Therefore,
   2 12 2 30
6 0
f   
  
Hence, there is local maximum at 2x  .

23. Evelyn
Further
   3 12 3 30
6 0
f   
 
Hence there is local minimum at 3x  .
Example 31: Find the points of local maxima and minima of the function
  5 4 3
5 5 1f x x x x   
Solution: The first derivative of the function is
 
 
  
4 3 2
2 2
2
5 20 15
5 4 3
5 1 3
f x x x x
x x x
x x x
   
  
  
Solve   0f x  for x
This gives
0, 1 and 3x x x  
Further
  3 2
20 60 30f x x x x   
If 1x  ,
       
3 2
1 20 1 60 1 30 1
10 0
f    
  
So, there is a local maximum at 1x  .
If 3x  ,
       
3 2
3 20 3 60 3 30 3
540 540 90
90 0
f    
  
 
So, there is a local maximum at 3x  .
If 0x  ,
       
3 2
0 20 0 60 0 300 0
0
f    

And the second derivative test fails. We now turn to first derivative test or general criteria for
local extrema to analyze what is happening at 0 . Let us apply the general criteria. We have
  2
60 120 30f x x x   
Therefore,
     
2
0 60 0 120 0 30
30
f    

Thus
   0 0 0f f   and  0 0f   .
Since 3n  , is odd therefore f has neither maximum nor minimum value at 0x  .

24. Evelyn
Example 32: Find all the points of local maxima and minima of the
function   4 3 2
8 22 24 1f x x x x x     .
Solution: The first derivative of the function is
 
   
3 2
4 24 44 24
4 1 2 3
f x x x x
x x x
    
   
Equate   0f x 
Hence,
   4 1 2 3 0x x x   
This implies that
1, 2 and 3x x x  
Further, the second derivative is
  2
12 48 44f x x x   
If 1x  ,
     
2
1 12 1 48 1 44
8 0
f    
 
So that, there is a local minima at 1x  .The corresponding minimum value is
         
4 3 2
1 1 8 1 22 1 24 1 1
8
f     
 
If 2x  ,
     
2
2 12 2 48 2 44
4 0
f    
  
So that, there is a local maxima at 1x  .The corresponding maximum value is
         
4 3 2
2 2 8 2 22 2 24 2 1
7
f     
 
If 3x  ,
     
2
3 12 3 48 3 44
8 0
f    
 
So that, there is a local minima at 1x  .The corresponding minimum value is
         
4 3 2
3 3 8 3 22 3 24 3 1
8
f     
 
Example 33: Find all the points of local maxima and minima of the function
   
2
1 ex
f x x  and also find the corresponding maximum and minimum values.
[C.A Foundation May 1994]
Solution: The first derivative of the function is

25. Evelyn
     
 
 
2
2
2
1 2 e +e 2 2
e 1 2 2 2
e 1
x x
x
x
f x x x x
x x x
x
    
    
 
That is
   2
e 1x
f x x  
For local maxima or minima equate   0f x  .
Hence,
 2
2
e 1 0
1 0 As e 0
1
x
x
x
x
x
 
  
 
Also, the second derivative is
   
 
2
2
e 1 2 e
e 2 1
x x
x
f x x x
x x
   
  
If 1x  ,
   1 2 1
1 e 1 1 2e
2e 0
f    
 
So, there is a local minimum at 1x  .
The corresponding minimum value is
   
2 1
1 1 1 e
0
f  

If 1x   ,
      
21 1
1 e 1 1 2 1 e
2
0
e
f 
      
  
So, there is a local maximum at 1x   .
The corresponding maximum value is
    
2 1
1 1 1 e
4
e
f 
   

Example 34: Find the points of local maxima and minima of the function
  3 2
3 9 15f x x x x   
Solution: The first derivative of the function is
 
  
2
3 6 9
3 1 3
f x x x
x x
   
  
Equate   0f x 

26. Evelyn
This implies that
1 and 3x x   .
If 1x   , then
    3 0;f x    
If 1 3x   , then     3 0;f x     and
If 3x  , then     3 0;f x    
Since  f x changes sign from  to  as x increases through 1 , therefore 1x   is a point of
local maximum. Similarly,  f x changes sign from  to  as x increases through3 ,
therefore 3x  is a point of local minimum.
Example 35: Find all the points of local maxima and minima of the function
4 3 2
8 22 24 1x x x x   
Solution: Differentiating the above function, to get
 
   
3 2
4 24 44 24
4 1 2 3
f x x x x
x x x
    
   
Equate   0f x  , to get
1, 2, 3x x x  
The second derivative of the function is
  2
12 48 44f x x x   
If 1x  , then the value of   8 0f x   , so f has a local minimum at 1x  . The local minimum
value of the function is  1 8f   .
If 2x  , then the value of   4 0f x    , so f has a local maximum at 2x  . The local
minimum value of the function is  2 7f  .
If 3x  , then the value of   8 0f x   , so f has a local minimum at 3x  . The local minimum
value of the function is  3 8f   .
Example 36: Find all the points of local maxima and local minima of the
function    2
1 ex
f x x  . Also find the function values at which these are maximum and
minimum. [C.A Foundation 2001]
Solution: Differentiating the above function we get
 
 
2
2
e e 2 e
1 e
x x x
x
f x x x
x
   
 
Now, to find local maximum and local minimum we equate the   0f x  . This implies that
 
2
1 e 0
1 as e 0
x
x
x
x
 
  
Again differentiating we get

27. Evelyn
     
2
1 e 2 1 ex x
f x x x    
If 1x   then the second derivative is   0f x  . Therefore the point 1x   is neither a local
maximum nor a local minimum point.
Example 37: The demand function for a particular commodity is 4
10e
x
p

 for0 6x  , where
p is the price per unit and x is the number of units demanded. Determine the price and quantity
for which total revenue is maximum.
Solution: Let R be the total revenue, then
  
4
price per unit number of units sold
10e
x
R
p x
x


 

To maximize the revenue R , differentiate R with respect to x :
4 4
4
1
10 e 1 e
4
10e 1
4
x x
x
dR
x
dx
x
 

  
    
  
 
  
 
For R to be maximum,
0
dR
dx

Thus,
4
10e 1 0
4
x
x  
  
 
This implies that
4x   since e >0x
Again differentiate
dR
dx
, to get
2
2
d R
dx
as
2
4 4
2
4 4
4
4
1
10 e 1 e
4
1 1 1
10e 10e
4 4 4
1
2.5e 1
4
3
2.5e
4
x x
x x
x
x
d R d
x
dx dx
x
x
x
 
 


   
     
    
      
          
      
  
     
  
  
    
  

28. Evelyn
When 4x  ,
2
4
2
11.875e 0
x
d R
dx

   , therefore R is maximum when 4x  . Substituting this
value of x in the demand function 4
10e
x
p

 , we get price
4
4
10e
10
e
p



And also we get the quantity for which the total revenue is maximum by substituting the value of
x in 4
10e
x
R x


Hence, the maximum revenue is
4
4
10e 4
40
e
R



Example 38: A TV company has 500 subscribers who are each paying Rs. 50 per month. The
company proposes to increase the monthly subscription and it is believed that for every increase
of Re 1, three subscribers will discontinue the service. Find what increase will yield maximum
revenue and what will this revenue be?
Solution: Let x be the increase in the monthly subscription in Rs . Then the new rate will
be50 x . Since for each increase of Re.1, 3 subscribers will discontinue the service. Thus, the
total number of subscribers left is500 3x . Let R denote the revenue of the company, the
  
  
2
2
rate number of subscribers
50 500 3
25000 150 500 3
25000 350 3
R
x x
x x x
x x

  
   
  
Now, differentiate above with respect to x , to get
350 6
dR
x
dx
 
Equate 0
dR
dx
 , to get
350 6 0
175
3
58.6
x
x
 


Also
2
2
6 0
d R
dx
  

29. Evelyn
Therefore, we conclude that R is maximum when the monthly subscription is increased by
Rs 58.6 and the maximum revenue is obtained by substituting
175
3
x  in
2
25000 350 3R x x  
2
2
25000 350 3
175 175
25000 350 3
3 3
25000 20416.66 10208.33
35,208
R x x  
   
     
   
  

Hence, the maximum revenue is 35,208
Example 39: A firm’s demand function is:
5
400logx
p
 
  
 
.Find the price and quantity where
total revenue is maximum. Also, find price elasticity of demand at that price.
Solution: Solving
5
400logx
p
 
  
 
for p , we obtain 400
5e
x
p
 
 
 
 and hence the total revenue R ,
is given by
400
5 e
x
R px x
 
 
 
 
Differentiating R , with respect to x , to obtain
400 400
400
1
5 e e 1
400
5e 1
400
x x
x
dR
x
dx
x
   
    
   
 
 
 
  
    
   
 
   
For R to be maximum, 0
dR
dx

That is,
1 0
400
400
x
x
 

400
Since 5e 0
x 
 
 
 
 
 
 
Also
2
400 400
2
400
1 1
5 e 1 e
400 400 400
1
e 2
80 400
x x
x
d R x
dx
x
   
    
   
 
 
 
      
          
       
 
   
 
At 400x  ,
2
1
2
1
e 0
80
d R
dx

  

30. Evelyn
Therefore R is maximum when 400x  . Substituting 400x  in 400
5e
x
p
 
 
 
 , to obtain price
5
e
p 
To evaluate elasticity of demand, we differentiate the demand function
 
5
400log 400 log5 logx p
p
 
   
 
With respect to p , to get
400dx
dp p
 
Therefore,
400
5
400log
1
5
log
400
d
p dx
x dp
p
p
p
p
x
  
 
   
   
 
 

 
 
 

Evaluating d at
5
e
p  or at 400x  , we obtain
1d 
Example 40: The manufacturing cost of an item consists of Rs 1000as overheads, material cost
Rs 2 per item and the labor cost
2
90
x
for x items produced. Find how many items are produced
to have the average cost as minimum.
Solution: The cost C is given by
2
1000 2
90
x
C x  
And the average cost is given by
2
1000 2
90
C
AC
x
x
x
x

 
  
 

31. Evelyn
1000
2
90
x
x
  
As we need minimum value of AC . We need to differentiate AC with respect to x and equate it
to0 . That is
 
2
1000
2
90
1000 1
90
d d x
AC
dx dx x
x
 
   
 
  
And
2
2
1000 1
0
90
90000
300
x
x
x
  


Also,
 
 
2
2 2
4
3
1000 1
90
1000
2
2000
0
d d
AC
dx dx x
x
x
x
 
   
 

 
Hence, AC is minimum when 300x  .
Example 41: If 2
0.01 5 100C x x   is a cost function, find the average cost function. At what
level of production x is there a minimum average cost? What is this minimum average cost?
Solution: The average cost function is
2
0.01 5 100
100
0.01 5
C
AC
x
x x
x
x
x

 

  
To minimize AC , we differentiate AC with respect to x , and equate it to zero
 
2
100
0.01 5
100
0.01
d d
AC x
dx dx x
x
 
   
 
 
And
2
100
0.01 0
x
 
2
10000
100
x
x



32. Evelyn
To determine if this value of x minimizes AC , we find the second derivative
 
 
2
2 2
4
3
100
0.01
100
2
200
0
d d
AC
dx dx x
x
x
x
 
  
 

 
Hence AC is minimum when 100x  and the minimum value of AC as by substituting
100x  in
100
0.01 5AC x
x
  
 
100
0.01 100 5
100
1 5 1
7
AC   
  

Example 42: A company produces x units of output at a total cost of 3 21
18 160
3
x x x 
Find
(i) The output at which marginal cost is minimum,
(ii) The output at which average cost is minimum, and
(iii) The output at which average cost is equal to marginal cost.
Solution: The total cost function is
3 21
18 160
3
C x x x  
(i) The marginal cost, MC is given by
2
36 160
dC
MC x x
dx
   
This implies that
2 36
dMC
x
dx
 
For MC to be minimum, 0
dMC
dx

That is,
2 36 0x  
18x 
Further,
2
2
2 0
d MC
dx
 
Thus, MC is minimum when 18x  .
(ii) The average cost, AC , is given by

33. Evelyn
21
18 160
3
C
AC x x
x
   
This implies that
2
18
3
dAC
x
dx
 
For AC to be minimum, equate 0
dAC
dx

That is,
2
18 0
3
27
x
x
 

Also,
2
2
2
0
3
d AC
dx
 
Thus, AC is minimum when 27x  .
(iii) We have to find the output of x so that AC MC
That is,
2 21
18 160 36 160
3
x x x x    
22
18 0
3
x x 
2
18 0
3
x x
 
  
 
This implies that
0 or 27x x  
Thus, the output at which average cost is equal to marginal cost is 27 units ( 0x  gives a trivial
solution).
Exercise 7.3:
1. Find the points of local maxima or minima, if any, of the following functions. Find also the
corresponding local maximum or local minimum values, as the case may be:
(i)   3 2
6 36 2f x x x x    (ii)   3 2
2 15 36 1f x x x x   
(iii)   2 3
8 36 3 2f x x x x    (iv)   3 2
2 9 12 1f x x x x    
2. Find the intervals on which the following functions are increasing or decreasing.
(i)   4 3 2
8 22 24 1f x x x x x     (ii)  
3
4
3
x
f x x 
(iii)      
3 3
1 3f x x x   (iv)   2
ex
f x x
3. Find the points of local maxima or minima, if any, of the following functions. Find also
corresponding local maximum or local minimum values, as the case may be
(i)   3 2
6 9 15f x x x x    (ii)   3 2
2 3 36 10f x x x x   

34. Evelyn
(iii)   3 2
2 9 12 25f x x x x    (iv)     
2
1 2f x x x  
4. Examine for maximum and minimum values the following functions:
(i)   4 3
3 4 1f x x x   (ii)   4 3 2
2 3 4 4f x x x x x    
5. Investigate the maximum and minimum values of the function
  5 4 3
12 45 40 6f x x x x   
6. Find the maximum and minimum values of the following functions:
(i)   2
1
1
x
f x
x x


 
(ii)  
2
7 6
10
x x
f x
x
 


7. Show that the function   x
f x x is minimum for -1
ex  .
8. Show that the function  
log x
f x
x
 has maximum value at ex  .
9. Show that the function   3 2
6 12 5f x x x x    has no maxima or minima.
10. Show that the function   3 2
9 30 5f x x x x    has no maxima or minima.
11. Show that the function   3 2
4 18 27 7f x x x x    has no maxima or minima.
12. Examine whether the function   3 2
3 3 7f x x x x    is
(i) maximum (ii) minimum (iii) neither maximum nor minimum
[C.A Foundation 2001]
Answers to Exercise 7.3:
1. (i) Increasing on  , 2  and  6, ; Decreasing on  2,6 .
(ii) Increasing on  ,2 and 2, ; decreasing on 2,3 .
(iii) Increasing on 2,3 ; Decreasing on  , 2  and  3, .
(iv) Increasing on 2, 1  ; Decreasing on  , 2  and  1,  .
2. (i) Increasing on  1,2 and  3, ; Decreasing on  ,1 and  2,3 .
(ii) Increasing on
1
,
4
 
 
 
; decreasing on
1
,
4
 
 
 
.
(iii) Increasing on 1, ; decreasing on ,1 .
(iv) Increasing on  , 2  and 0, ; Decreasing on  2,0 .
3. (i) Local maximum at 1x  and value 19 ; Local minimum at 3x  and value 15
(ii) Local maximum at 3x   and value 91 ; Local minimum at 2x  and value 34 
(iii) Local maximum at 1x  and value 30 ; Local minimum at 2x  and value 29
(iv) Local maximum at 2x   and value 0 ; Local minimum at 0x  and value 4 
4. (i) Local maximum at 1x  and value 0 ; point of inflection at 0x 
(ii) Local maximum at
1
2
x   and
81
value
16
 ; Local minimum at 2x  and value 0 ; and
Local minimum at 1x  and value 0

35. Evelyn
5. Local maximum at 1x  and value 13 ; local minimum at 2x  and value 10  ; point of
inflection at 0x 
6. (i) Maximum
1
value
5
 and occurs at 2x  ; Minimum value 1 and occurs at 0x  .
(ii) Maximum value 1 and occurs at 4x  ; Minimum value 25 and occurs at 16x  .
12. Neither maximum nor minimum
7.4 Concavity and Convexity
Let us consider  f x to be a differentiable function in the interval ,a b . The let us assume that
 f x exists and is less than0 , that is   0f x  than it is said to be concave and if  f x exists
and is greater than0 , that is   0f x  than it is said to be convex.
7.4.1 Conditions for concavity and convexity
The following are the conditions for concavity and convexity
(i) For concavity
A function  f x should be differentiable in the interval  ,a b and  f x exists and the value of
it is   0f x  .
(ii) For convexity
A function  f x should be differentiable in the interval  ,a b and  f x exists and the value of
it is   0f x  .
7.4.2 Inflection point:
An inflection point is a point on the curve where the concavity changes sign from plus to minus
or minus to plus.
Let us consider  f x to be a differentiable function in the interval ,a b . The let us assume that
 f x exists, the point at which   0f x  or the point at which  f x does not exist is called as
inflection point.
7.4.3 Condition for inflection point:
The condition for inflection point to exist is that the function  f x is differentiable in the
interval  ,a b and the point of inflection c at which   0f x  satisfies   0f c 

36. Evelyn
Example 43: A company, manufacturing washing machines, charges Rs 6000per machine for
an order of 50 machines or less. For each order in excess of50, the company reduces the charge
by Rs 50. Find the largest size order to maximize the total revenue.
Solution: Let x number of washing machines is the size of order to maximize total revenue.
Then, revenue function is
   
 
2
6000 50 50
6000 50 2500
7500 50
R x x x
x x
x x
    
  
 
For  R x to be maximum,   0R x  and   0R x 
Now,
  7500 100 0
75
R x x
x
   

And
  100 0R x   
Thus, the revenue  R x is maximum at 75x  . So, when 75washing machines are ordered then
the total revenue is maximum.
Example 44: A company has for x items produced the total cost  C and total revenue  R
given by equations 2
100 0.15C x  and 3R x . Find how many items are produced to maximize
the profit and what is the profit? [Delhi University B.Com 2008]
Solution: Let P denote the profit function. Then
2
3 100 0.15
3 0.03
P R C x x
dP
x
dx
    
 
To maximize profit we set 0
dP
dx
 and solve for x , thus we get
3 0.03 0
100
x
x
 

Also,
2
2
0.03 0
d P
dx
  
Thus, profit is maximum when 100x  units. The maximum profit is obtained by putting
100x  in profit function. Thus, maximum profit obtained is
300 100 150 50P    
Example 45: A company has for x items produced the total cost C and total revenue R given
by equations 2
100 0.01C x  and 2R x . Find how many items are produced to maximize the
profit. What is this profit?
Solution: Let denote the profit function.
Then

37. Evelyn
 2
2
2 100 0.01
2 100 0.01
2 0.02
P R C
x x
x x
dP
x
dx
 
  
  
 
To maximize profit, we set 0
dP
dx
 and solve for x .
Thus
2 0.02 0
100
x
x
 

Also,
2
2
0.02 0
d P
dx
  
Thus profit P is maximum when 100x  and the maximum profit is obtained by substituting
100x  in profit function.
We get
 
2
Maximum Profit 200 100 0.01 100
200 100 100
0
  
  

Exercise 7.4:
1. The demand equation for a manufacture’s product is
80
4
x
p

 where x is the number of units
and p is price per unit. At what value of x will there be maximum revenue? What is the
maximum revenue?
2. For a monopolist’s product, the demand function is 0.02
10,000e p
x 
 . Find the value of p for
which the maximum revenue is obtained.
3. A company charges Rs 15000 for a television set on order of 60 or less sets. The charge is
reduced on every set by Rs 100per set for each ordered in excess of 60. Find the largest size
order the company should allow so as to receive maximum revenue.
4. A tour operator charges Rs 136 per passenger for 100 passengers with a refund of Rs 4 for
each 10 passengers in excess of 100. Determine the number of passengers that will maximize the
amount of money the tour operator receives.
5. The owner of a 40 unit model can rent all units nightly at Rs 1000 per night. However per
each Rs 100per night increase in rent, 2 units will be vacant with no possibility of filling them.
How many units should be rent per night and at what rate to maximize his daily income?
6. A steel plant is capable of producing x tons per day of low- grade steel and y tons per day of
high-grade steel, where
40 5
10
x
y
x



. If the fixed market price of low-grade steel is half of that of

38. Evelyn
high- grade steel, show that about
1
5
2
tons of low-grade steel are produced per day for maximum
revenue.
7. A manufactures total cost function is given by
2
3 400
4
x
C x   where x is the number of
units produced. At what level of output will the total cost be a minimum? What is this minimum?
8. Let the cost function of a firm be given by 2 31
300 10
3
C x x x   , where stands for cost and
for x output. Calculate
(i) The output at which marginal cost is minimum.
(ii) The output at which average cost is minimum.
(iii) The output at which average cost is equal to marginal cost.
9. If x be the number of workers employed, the average cost of production is given by
 
3
24
2 4
AC x
x
 

Show that the value 4.25x  will make the expression minimum. In the intent management will
you then advise to employ four or five workers? Give reasons for your answer.
10. A firm produces x units of output per week at a total cost of 3 21
Rs 5 3
3
x x x
 
   
 
. Find the
output levels at which the marginal cost and the average variable cost attain their respective
minima.
11. A machine initially costs Rs 6400 with no scrap value. The cost of operating is Rs 500in the
first year and increases by Rs 800 in each successive year. Determine
(i) The number of years it be operated for minimizing total cost per year, and
(ii) Corresponding cost per year.
12. The cost of fuel in running a vehicle is proportional to the square of the speed and is
Rs 45per hour for a speed of 15 km per hour. Other expenses amount to Rs 500 per hour. Find
the most economical speed for the vehicle and also the minimum cost for a distance of 100 km.
13. The cost of fuel running a train is proportional to cube of the speed is 12 km. per hour, and is
Rs 64per hour when the speed is km. Other costs amount to Rs 2000 per hour. Find the most
economical speed of the train in running a distance of 100 km.
14. Show using calculus that profit of a monopolist is maximum when
(i) MR MC ; and
(ii) Slope of Slope ofMR MC . [Delhi University B.Com 2008]
15. A firm has found from past experience that its profit in terms of number of units x produced
is given by
 
3
729 2500
3
x
P x x    ,0 35x  .
Compute
(i) The value of x that maximizes the profit,
(ii) The profit per unit of the product when this maximum level is achieved.

39. Evelyn
16. A sitar manufacturer can sell x sitars per week at Rs p each. The cost of production
is 21
Rs 500 13
5
x x
 
  
 
. Find how many sitars should be manufactured for maximum profit and
what is this profit?
17. A radio manufacturer produces x sets per week at a total cost of  2
Rs 78 2000x x  . He is a
monopolist and the demand function for his product is
 600
8
p
x

 , when the price is p per set.
Show that maximum net revenue (that is profit) is obtained when 29 sets are produced per week.
What is the monopoly price?
18. A radio manufacturer produces x sets per week at a total cost of
2
Rs 3 100
25
x
x
 
  
 
. He is a
monopolist and the demand for his product is 75 3x p  where p is the price in rupee per set,
Show that the maximum net revenue is obtained when 30 sets are produced per week. What is
the monopoly price? [Delhi University B.Com (H) 2008]
19. If the total cost of a firm is   3 21
5 30 10
3
C x x x x    where C is the total cost and x is the
output, and price under perfect competition is given as 6, find for what value of x the profit will
be maximized. Examine both first and second order conditions.
20. Find the profit maximizing output level, given 240 10x p  and 10
25
x
AC   , where x
represents the units of output, p represents price, and AC represents average cost.
21. The production function of a commodity is given by
2 31
40 3
3
Q F F F   ,
Where Q is the total out and F is the input
(i) Find the number of units of input required to give the maximum out put
(ii) Find the maximum value of marginal product
(iii) Verify that when the average product is maximum, it is equal to marginal product.
22. The product function for a commodity is given by
2 3
Q aF bF cF   ,
Where Q is output and F is input
Verify that when average product is maximum, it is equal to marginal product. Are there any
restrictions for values of , ,a b c ?
23. The relationship between sales,  TR x and advertising cost Rs x , is given by
 
32,000
500
x
TR x
x


It is known that gross profit is 25 percent of the sales. Determine
(i) The corresponding net profit as a function of  ,x P x
(ii) The value x of which maximizes  P x , and
(iii) Maximized  P x

40. Evelyn
24. A manufactures of a product finds that for the first 500 units that are produced and sold, the
profit is Rs 50 per unit. The profit on each of the units beyond 500 is decreased by Rs 0.10
times the number of additional units produced. What level of output will maximize profit?
Answers to Exercise 7.4:
1. 40,400
2. 50
3. 105
4. 220
5. 30 Units per night, Rs. 1,500
7. 40, minAC 23x  
8. (i) 10 (ii) 15 (iii) 15
9. 5Workers
10.
3
1 and
2
11. (i) 4years (ii) Rs. 3,300
12. 50kms/hour, Rs. 2,000
13. 30kms/hour
15. (i) 27 (ii) Rs. 393.40
16. 30, max. profit Rs. 1,180x  
17. Rs. 368
18. Rs. 15
19. 6x 
20. 50
21. (i) 10 (ii) 49
22. b positive, c negative
23. (i)  
8000
500
x
P x x
x
 

(ii) Rs. 1,500 (iii) Rs. 4,500
24. 75x 
LIST OF IMPORTANT FORMULAE
1.
   
0
lim
h
x c
f c h f cdy
dx h

 

2. If c is a constant then   0
d
c
dx

3. If n is any real number then   1n nd
x nx
dx


4. If f is a differentiable function and c is a constant then
   .
d d
c f x c f x
dx dx
      
5. If f , g are differentiable functions and c is a constant then

41. Evelyn
       
d d d
f x g x f x g x
dx dx dx
            
6. If f , g are differentiable functions then
           
d d d
f x g x f x g x g x f x
dx dx dx
           
7. If f , g are differentiable functions then
 
 
       
 
2
d d
g x f x f x g xf xd dx dx
dx g x g x
       
 
    
8. If  y f u is a differentiable function in u and  u g x is a differentiable function in x , then
 y f g x    is a differentiable function in x and
dy dy du
dx du dx

9. Let n
y u and  u f x .
     
1n nd
f x n f x f x
dx

       
10.  
1
log loga a
d
x x
dx x

11.  e
1
log
d
x
dx x

12.   logex x ad
a a
dx

13.  e ex xd
dx

14. If eu
y  then eudy du
dx dx

15. If the position at time t of an object moving along a straight line is given by  s t , then the
object has velocity is represented as rate of change of  s t and it is given as    
ds
v t s t
dt
  and
acceleration is represented as rate of change of velocity  v t and it is given
as      
2
2
dv d s
a t v t s t
dt dt
     .

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7.5 Summary:
In calculus the concepts of maxima and minima are very useful in the field of business
economics which help us to solve the problems related to them and also gives us solutions
regarding them. Otherwise it would be a difficult process to find the same. The concept of
concavity and convexity is also useful in determining the curve nature and to say whether the
curve has maxima or local minima. In this chapter we have study the conditions related to local
maxima, local minima, how to find out them and also the concavity, convexity and there
conditions and know about the inflection point.
7.6 Key Terms:
Derivative: The derivative of  y f x , can be written as  
dy
f x
dx
 . It represents the rate of
change in the function with respect to x .
Differentiation: The method of finding the derivative of a function is called differentiation.
Local maxima: The function  f x is said to have a local maximum at 0x x if    0f x f x for
all values of x sufficiently close to 0x .
Local minima: The function  f x is said to have a local minimum at 0x x if    0f x f x for
all values of x sufficiently close to 0x .
Concavity: This is the condition when  f x exists and is less than0 , that is   0f x  and the
function is said to be concave.
Convexity: This is the condition when  f x exists and is less than0 , that is   0f x  and the
function is said to be concave.
Inflection point: An inflection point is a point on the curve where the concavity changes sign
from plus to minus or minus to plus.
Stationary point: A point 0x in the domain f of at which  0 0f x  is called a stationary point
of f .
7.7 Questions and Exercise
Short Answer Questions
1. A manufacture determines that when x thousand units of a particular commodity are produced
the profit generated will be   2
400 6800 12000P x x x    dollars. At what rate is profit
changing with respect to level of production x when 9000 units are produced?
2. It is estimated that x months from now, the population of certain community will be
  2
20 8000P x x x  
a. At what rate will the population be changing with respect to time 15months from now?
b. By how much will the population change during the 16th month?
3. If   4
f x x then find
   f x h f x
h
 
?

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4. Differentiate     1 3 2P x x x   using product rule?
5. Differentiate  
2
5 7
2
x x
Q x
x
 
 using quotient rule?
6. Differentiate the function 2
2 4 1
3 3 5
x x
y
x x

    ?
7. Find the second derivative of the function   4 2
5 3 3 7f x x x x    ?
8. Find the second derivative of the function  2
3 1y x x  ?
9. If the position of an object moving along a straight line is given by   3 2
3 4s t t t t   at
timet , find its velocity and acceleration.
10. The cost of producing units of a particular commodity is   21
4 53
3
C x x x   dollars, and
the population level t hours into a particular production run is   2
0.2 0.03x t t t  units. At what
rate is cost changing with respect to time after 4 hours.
11. Differentiate the function   2
3 2f x x x   .
12. Differentiate the function    
34
2f x x x  .
13. Find the second derivative of the function  
 
2
3 2
1
x
f x
x



.
Long Answer Questions
14. The manager of an appliance manufacturing firm determines that when blenders are priced at
p dollars per piece, the number sold each month can be modeled by
 
8000
D p
p

The manager estimates that t months from now, the unit price of the blenders will be
 
3
2
0.06 22.5p t t  dollars. At what rate will the monthly demand for blenders  D p be
changing 25 months from now? Will it be increasing or decreasing at this time?
15. A manufacture of digital cameras estimates that when x hundred cameras are produced, the
total profit will be   3 2
0.0035 0.07 25 200P x x x x     thousand dollars.
(a) Find the marginal profit function.
(b) What is the marginal profit when the level of productions are 10x  , 50x  and 80x  ?
16. Find the derivative  y f x of the following
(a)   3 2
7 2f x x x  
(b)     2 2
5 3 2f x x x x   
17. The manager of a company determines that when q hundred units of particular commodity
are produced, the total cost of production is C thousand dollars, where 2 3
4 4275C q  . When

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1500 units are being produced, the level of production is increasing at the rate of 20 units per
week. What is the total cost at this time and at what rate is it changing?
Answers to Short Answer Types Questions
1. 400 dollars per thousand units.
2. a. 50people per month
b. 51 people
3. 3 2 2 3
4 6 4x x h xh h  
4. 6 5x 
5. 2
1 7
2 2x

6. 3 2
4 1 1
3 3x x
  
7. 2
60 6x 
8. 18 2x 
9. 2
velocity 3 6 4,acceleration 6 6t t t    
10. $10.13 per hour
11.
2
2 3
2 3 2
x
x x

 
12.    
24 3
3 2 8 1x x x 
13.
 
4
6
1
x
x 
Answers to Long Answer Types Questions
14. 4 units per month, decreasing
15. (a) 2
0.0105 0.14 25x x  
(b) 25.35,5.75, 31
16. (a) 2
3 14x x (b)     2 2
5 1 4 2 3 2x x x x x       
17.120,000 thousand dollars , increasing at the rate of $1687.50 per week.