1) The document discusses finding the maximum and minimum values of functions with constrained variables. It provides definitions and a working rule for determining extreme values in multiple steps. 2) Examples are provided to demonstrate the process of finding the minimum/maximum value and stationary points of functions subject to constraints. 3) The final example solves for the points on a surface nearest to the origin and the minimum distance.

1. Maxima & Minima with
Constrained Variables
BY:
Arpit Modh (16BCH035)
B.Tech Chemical
Nirma University,
Ahmedabad.

2. Definitions:-
Let, u = f (x , y) be a continuous function of x and y. Then u
will be maximum at x = a, y = b, if f (a ,b ) > f(a + h , b + k)
and will be minimum at x=a, x=b, if f(a, b) < f(a + h, b + k)
for small positive or negative values of h and k.
 The point at which function f(x, y) is either maximum or
minimum is known as stationary point.
 The value of the function at stationary point is known
as extreme (maximum and minimum) value of function
f(x, y).

3. Working Rule:-
To determine the maxima and minima (extreme values) of a
function f(x, y).
Step 1: Solve ∂f/ ∂x = 0 and ∂f/ ∂y = 0 simultaneously for x
and y.
Step 2: Obtain the values of r= ∂²f/ ∂x², s= ∂²f / ∂x²,
t= ∂²f/ ∂x².

4. Step 3:
(i) If rt - s² > 0 and r < 0 (or t < 0) at (a, b) then f(x, y) is
maximum at (a, b) and the maximum value of the
function is f(a, b).
(ii) If rt - s² > 0 and r > 0 (or t > 0) at (a, b) then f(x, y) is
minimum value of the function is f(a, b).
(iii) If rt - s² < 0 at (a, b) then f(x, y) is either maximum nor
minimum at (a, b). Such a point is known as saddle point.
(iv) If rt - s² = 0 at (a, b) then no conclusion can be made
about the extreme values of f(x, y) and further
investigation is required.

5. Example 1
Find the minimum value of x² + y² + z² with the
constraint x + y + z = 3a.
Solution: f = x² + y² + z²
x + y + z = 3a
z = 3a - x - y …..(1)
substituting the value of z in Eq. (1),
f = x² + y² + (3a –x- y) ²
Step 1 For extreme values,
∂f/ ∂x = 0 and ∂f/ ∂y = 0
2x – 2(3a - x - y ) = 0 2y - 2(3a - x - y) = 0
4x - 6a + 2y = 0 2y - 6a + 2x + 2y = 0
2x + y = 3a x + 2y = 3a
……(2) ….(3)

6. Solving Eqs (2) and (3),
x = y = a
The stationary point is (a, a).
Step 2 r = ∂²f/ ∂x² = 4
s = ∂²f/ ∂x ∂y = 2
t = ∂²f/ ∂y² = 4
Step 3 At (a, a), r = 4, s = 2, t = 4
rt - s² = (4)(4) – (2) ² = 12 > 0
Also, r = 4 > 0
Hence, f(x, y) is minimum at (a, a)
fmin = a² + a² + (3a - a - a) ² = 3a²

7. Example 2
Divide 120 into three parts so that the sum of their products taken
two at a time shall be maximum.
Solution: Let x, y, z be three numbers.
x + y + z = 120
f = xy + yz + xz
= xy + y(120 - x - y) + x( 120 - x - y)
= xy + 120y – xy - y² + 120x - x² -xy
= 120x + 120y - xy - x² - y²
For extreme values, ∂f/∂x = 0
120 - y - 2x = 0 …(1)
And ∂f/ ∂y = 0
120 - x - 2y = 0 ….(2)
Solving Eqs (1) and (2),
x = 40
y = 40
Stationary point is (40, 40).

8. Step 2 r = ∂f/ ∂x = -2
s = ∂f/ ∂x ∂y = -1
t = ∂f/ ∂y = -2
Step 3 At (40, 40)
rt - s² = (-2)(-2) – (-1) ² = 3 > 0
r = -2 < 0
Hence, f(x, y) is maximum at (40, 40).

9. Example 3
Find the points on the surface z²=xy+1 nearest to the origin. Also find
that distance.
Solution:
Let p(x, y, z) be any point on the surface z² = xy + 1.
Its distance from the origin is given by
d² = x² + y² + z²
Since p lines on the surface z² = xy + 1
d² = x² + y² + xy + 1
Let f(x, y) = x² + y² + xy + 1
Step 1 For extreme values ,
∂f/ ∂x=0
2x+y=o
∂f/ ∂y=0
2y+x=0
Solving Eqs (1) and (2),
x=0 , y=0

10. Step 2
r = ∂²f/ ∂x² = 2
s = ∂²f/ ∂x ∂y = 1
t = ∂² f/ ∂y² = 2
Step 3 At (0,0), r = 2, t = 2, s = 1
rt - s² = (2)(2) - 1² = 3 > 0
Also, r = 2 > 0
f(x, y) , i.e. d ² is minimum at (0,0) and hence d is minimum at (0,0).
At (0,0),
z²=xy+1=1
z=+1,z=-1
Hence, d is minimum at (0,0,1) and (0,0,-1).
The points (0,0,1) and (0,0,-1) on the surface z² = xy + 1 are the
nearest to the origin Minimum distance = 1.