©
Alex Teshon
Daffy Durairaj
The term optimization means to optimize
something, or use something at its best.
This refers, both in real life and in
Calculus, to the maximum or minimum
value of something.
 When you optimize, you try to find the
maximum or minimum value required for
the given problem

What is Optimization Anyways?
L
BOX:
LWH
H
W
CONE:
⅓πr²h
h
r
h
CYLIND
ER:
πr²h
r
SPHERE:
¾πr³
r
Surface
Area
Area
Cylinder
without top:
πr²+2πrh
Circle:
πr²
w
Rectangle:
2l+2w
Cylinder
with a closed
top:
2πr²+2πrh
l
Steps of Optimization
 BEFORE YOU DO ANYTHING MAKE SURE YOU
READ THE PROBLEM CAREFULLY!!!!
1.DRAW THE PICTURE
2.LABEL THE PICTURE WITH YOUR X’S AND Y’S
3. WRITE OUT ANY EQUATIONS YOU WILL
NEED
3. SOLVE FOR EITHER Y OR X
5. AFTER SOLVING FOR YOUR VARIABLE, PLUG
THE SOLVED VARIABLE BACK INTO THE
ORIGINAL EQUATION
Even More Steps
6. TAKE THE DERIVATIVE OF THAT EQUATION
7. SET THIS DERIVATIVE EQUAL TO ZERO
8. SOLVE FOR EITHER Y OR X
9. TAKE THE SECOND DERIVATIVE TO DETERMINE
IF THIS IS THE MAX OR MIN
10. PLUG THE SOLVED FOR VARIABLE BACK INTO
THE EQUATION TO SOLVE FOR THE OTHER
VARIABLE
11.PLUG BOTH THE SOLVED VARIABLES BACK INTO
THE EQUATION TO GET THE FINAL ANSWER
Lets Try This Out!
 A shepherd wishes to build a rectangular fenced area
against the side of a barn. He has 360 feet of fencing
material, and only needs to use it on three sides of the
enclosure, since the wall of the barn will provide the last
side. What dimensions should the shepherd choose to
maximize the area of the enclosure?
DRAW THE PICTURE!!!
x
y
y
BARN
Use the needed equations to solve for
either x or y
Since we are dealing
with area and
perimeter we only
need 2 equations:
A= xy
P= 2y+x
We know that the
perimeter has to be 360
feet so we can plug this
into the perimeter
equation
P= 360
360= 2y+x
In this case, we will
solve for x
360= 2y+x
X= 360-2y
Plug it back in and get the derivative
Now that we have solved for x, we can plug this into the other equation
to solve for our other variable
A= xy
X= 360-2y
A=(360-2y)y
A= 360y-2y²
Now that we have our equation,
take the derivative
A= 360y-2y²
A’=360-2y
SET THE DERIVATIVE EQUAL TO ZERO
A’= 360-4y
4(90-y) = 0
y = 90 ft
Now that we have our y value, we can plug this back into the
original equation to get our answer!
360= 2y+x
Y= 90
360= 2(90)+x
360=180+x
X= 180
So to get the maximum are
of the fence, y should be
90 ft and x should be 180
ft.
Take the derivative again
 Once you find your value to plug in, take the second
derivative to find if what you have is a relative maximum or
minimum
 A’= 360-4y
A’’= -4
if A’’<0 then the value is a relative max
If A’’>0 then the value is a relative min
-4<0
Therefore this is
a relative max
PROBLEM #1

An open rectangular box with square base is to
be made from 48 ft.2 of material. What
dimensions will result in a box with the largest
possible volume ?
DRAW IT OUT!!!
SA:
X²+4XY
V: X²Y
y
x
x
SA= x²+4xy
SA= 48
48= x²+4xy
V= x²y
4xy= 48-x²
Y= 48-x²/4xy
V= 12x²/x – ¼ x³
V= 12x- ¼x³
Y= 12/x – ¼x
V’= 12- ¼ x²
¾(16-x²)
¾(x-4)(x+4)
V= x²(12/x – ¼x)
X= 4 x≠0, -4
Y= 12/(4) – (4)/4
3-1
Y= 2
V= x²y
V= (4)²(2)
V= 32 ft³
PROBLEM #2

A container in the shape of a right circular
cylinder with no top has surface area 3 ft.2
What height h and base radius r will maximize
the volume of the cylinder ?
DRAW IT OUT!!!
r
h
SA:
πr²+2πrh
V: πr²h
SA= πr²+2πrh
SA= 3π
3π= πr²+2πrh
2πrh= 3π- πr²
h= 3π- πr²/2πr
3/2r- ½r
V= πr²h
V= πr²(3r/2 – ½r)
V= 3πr²/2r - πr³/2
3πr/2- πr³/2
h= 3/2r- ½r
h= 3/2(1) – ½(1)
3/2 – 1/2
h= 1
V’= 3π/2 - 3πr²/2
3π/2(1-r²)
3π/2(1-r)(1+r)
r= 1
r≠ 0,-1
V= π(1)²(1)
V= π ft³
PROBLEM #3

A piece of sheet metal is rectangular, 5ft wide
and 8ft long. Congruent squares are to be cut
from its corners. The resulting piece of metal is
to be folded and welded to form an open top
box. How should this be done to get a box of
largest possible volume?
DRAW IT OUT!!!
8
8-2x
5
x
x
5-2x
x
8-2x
5-2x
V=lwh
V= x(8-2x)(5-2x)
(8x-2x²)(5-2x)
4x³-26x²-40x
6
1
1
3
V’=12x²-52x-40
4(3x²-13x-10)
4(3x-10)(x-1)
X= 1, 10/3
X≠ 10/3
1
X= 1
5-2(1)= 3
8-2(1)= 6
6
The cut outs will be 1 in on each side and the lengths of the box will
be 6x3x1
3
 Find
the maximum volume of a right cylinder
that can be inscribed in a cone of altitude 12
inches and a base radius 4 inches if the axes
of the cylinder and cone coincide
A
rectangular plot of land containing 216
square meters is to be enclosed by a fence
and divided into two equal parts by another
fence parallel to one of the sides. What
dimensions of the outer rectangle require
the smallest total length for the two fences?
A
printed page has 1 inch margins at the top
and .75 on the sides. If the area of the
printed paper is to be 48 square inches, what
should the dimensions be?



A window is in the shape of a rectangle surmounted by a
semicircle. Find the dimensions when the perimeter is 24 meters
and the area is as large as possible
A container with a rectangular base, rectangular sides, and no top
is to have a volume of 2 cubic meters. The width of the base is to
be 1 meter. When cut to size, material costs $10 per square meter
for the base and $5 per square meter for the sides. What is the
cost of the least expensive container?
A circular cylindrical container, open at the top and having a
capacity of 24π cubic inches, is to be manufactured. If the cost of
the material used for the bottom of the container is three times
that used for the curved part and there is no waste of material,
find the dimensions which will minimize the cost.
Getting Ready For The AP Exam
1979 AB3 BC3
Find the maximum volume of a box
that can be made by cutting out
squares from the corners of an 8
inch by 15 inch rectangular sheet
of cardboard and folding up the
sides.
Let’s See How You Did
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http://www.sparknotes.com/math/calcab/ap
plicationsofthederivative/problems_8.html
http://www.qcalculus.com/cal08.htm
http://www.math.ucdavis.edu/~kouba/CalcO
neDIRECTORY/maxmindirectory/MaxMin.html