The document provides an example problem involving calculus applications to find the local extrema and absolute maximum and minimum of a function f(x) = x^3 - 45x^2 + 600x + 20 on the interval [0,30]. It shows the steps to take the derivative of the function, set it equal to 0 to find critical points, and use the derivative test to determine if the critical points are maxima or minima. The document then evaluates the function at the endpoints of 0 and 30 to determine the absolute maximum and minimum values on the given interval.

1. Calculus Applications
Math Studies 1

2. f ( x) = x − 45 x + 600 x + 20
3 2
a) Find the local extrema and identify them
as either a local maximum or a local
minimum.
b) Find the coordinates of the absolute
maximum and absolute minimum of the
function in the interval [ 0,30]

3. First,
f ' ( x) = 3 x − 90 x + 600
2
differentiate.
0 = 3 x − 90 x + 600
2
Set f’(x) to 0
0 = x − 30 x + 200
2
Solve for x
0 = ( x − 10)( x − 20) Factor
Hence, we have local extrema at x = 10 and x = 20

4. To identify them as either maxima or minima, we can use
the derivative -
f ' (9) = 3(9) 2 − 90(9) + 600 = 33
f ' (11) = 3(11) − 90(11) + 600 = −27
3

5. To identify them as either maxima or minima, we can use
the derivative -
f ' (19) = 3(19) 2 − 90(19) + 600 = −27
f ' (21) = 3(21) − 90(21) + 600 = 33
3

6. b) Check the endpoints, x = 0 and x = 30
f (0) = (0) 3 − 45(0) 2 + 600(0) + 20 = 20
f (30) = (30) − 45(30) + 600(30) + 20 = 4520
3 2

7. A rectangular pen is to be fenced in using two types of
ncing. Two opposite sides will use heavy duty fencing at $3/ft
hile the remaining two sides will use standard fencing at $1/ft.
What are the dimensions of the rectangular plot of greatest area
at can be fenced in at a total cost of $3600?
A = xy 2 ( 3x ) + 2 ( 1y ) = 3600
3x + y = 1800
1y y = 1800-3x
A = x ( 1800 − 3x )
3x A = 1800x − 3x 2
A ' = 1800 − 6x A " = −6
0 = 1800 − 6x Therefore max
300 = x
900 = y
The dimensions of a rectangular plot of greatest area are 300 x 900

8. 4. An open-top box with a square bottom and rectangular sides
is to have a volume of 256 cubic inches. Find the dimensions
that require the minimum amount of material.
x
S = x + 4xy
2
 256 
→ S = x + 4x  2 
2
y
V = x y = 256
2
 x  x
1024
y = 256/x² S = x2 +
x
1024 2048
S' = 2x − 2 S" = 2 + 3 > 0
x x
1024 therefore a min
0 = 2x − 2
x
x=8 →y=4
8x8x4

9. Solve this problem.

10. Solve this problem.