AP Calculus Review:
Optimization
Created by Sarah Tuggle
Basic Steps to Solve
Optimization Problems
1. Read the problem
2. Draw a picture to help you visualize
the situation
3. Write 1 equation for each variable.
(You should have the same number
of equations as variables)
Basic Steps to Solve
Optimization Problems
4. Take the equation with the number
so you can solve for one of the
variables
5. Substitute the variable you found into
one of the other equations
6. Take the derivative and set the
equation equal to 0
Basic Steps to Solve
Optimization Problems
7. Take the second derivative to check
your answer (if it’s <0 max, >0 min)
8. Solve for the other variables needed
to answer the question
Example Problem
1. Read the Problem
Find the area of the largest rectangle
with a perimeter of 200 feet.
2. Draw a picture that represents the
scenario
y
x
3. Write an equation for each variable.
There are two variables therefore
there should be two equations.
A= x+ y
200= 2x+ 2y
4. Take the equation with a number in it
and solve for one of the variables.
100= x+ y
x= 100-y
5. Substitute that variable into the other
equation and solve for the other
variable.
A= xy x= (100-y)
A= (100-y)y
A= 100y -y²
6. Take the derivative and set it equal to
0 to solve for the variable
A′= 100- 2y
0= 100-2y
y= 50
7. Take the second derivative to check
your answer (<0 max, >0 min)
A′′= -2 < 0 max
8. Solve for the other variables
needed to answer the question
200= 2x+ 100
x= 50
A= 50(50)
A= 2500 sq ft
Try This!!!
A tank with a rectangular base and
rectangular sides is to be open at the
top. It is to be constructed so that it’s
width is 4 meters and volume is 36
cubic meters. If building the tank cost
$10 per sq meter for the base and $5
per sq meter for the sides, what is the
cost of the least expensive tank?
Draw a picture
x
4
y
Write an equation for each variable
36= 4xy
C= 10(4x)+ 5(8y)+ 5(2xy)
C= 40x+ 40y+ 10xy
Take the equation with the variable
and solve for one of the variables
36= 4xy
9= xy
y= 9
x
Substitute the variable you solved
for into the other equation
C= 40x+ 40(9)+ 10x(9)
x
x
C= 40x+ 360+ 90
x
Take the derivative and set it equal
to 0 to solve for the variable
C′= 40- 360/x²
0=40- 360/x²
40= 360/x²
40x²= 360
x²= 9
x= 3
Take the second derivative (< 0
max, > 0 min)
C′′= 720/x³ > 0 min
Solve for the other variables
needed to answer the question
x= 3
36= 4(3)y
y= 3
C= 40(3)+ 40(3)+ 10(3)(3)
cost= $330
Here’s another problem to try!
A closed box with a square base is to have a
volume of 2000 cubic in. The material for
the top and bottom of the box is to cost $3
per sq in, and the material for the sides is to
cost $1.50 per sq in. If the cost of the
material is to be the least, find the
dimensions of the box.
Draw a picture
x
x
y
Write an equation for each
variable
2000= x²y
A= 3(2x²)+ 1.5(4xy)
A= 6x²+ 6xy
Take the equation with the number
and solve for one of the variables
2000= x²y
y= 2000
x²
Substitute the variable you solved
for into the other equation
A= 6x²+ 6x(2000)
x²
A= 6x²+ 12000
x
Take the derivative and set it equal to
0 to solve for the variable
A′= 12x- 12000/x²
0= 12x- 12000/x²
12x³=12000
x³=1000
x=10
Take the second derivative (<0
max, >0 min)
A′′= 12+ 24000 >0 min
x³
Solve for the other variables
needed to answer the question
y= 2000/10²
y= 20
Dimensions: 10in x 10in x 20in
Here’s something a little more difficult
A window is in the shape of a
rectangle surmounted by a
semicircle. Find the dimensions
when the perimeter is 12 meters
and the area is as large as
possible.
12= 2y + x+ ½πx
A=xy+ ⅛πx²
12- x- ½πx= 2y
6- ½x- ¼πx= y
A=x(6- ½x- ¼πx)+ ⅛πx²
A=6x- ½x²- ⅛πx²
A′=6- x- ¼πx
0=6- x- ¼πx
6=x+ ¼πx
6=x(1+ ¼π)
x= 24/(4+π)
A′′-1- ¼π <0 max
y=6- 12/(4+π)- 6π/(4+π)
y= 12/(4+π)
Bibliography
• Youse, Bevan K., and F. Lane Hardy.
Calculus with Analytical Geometry .
1968. Reprint. New York: Holt,
Rinehart and Winston, 1978. Print.
© Sarah Tuggle 2011