OPTIMIZATION PROBLEMS
AP Calculus AB
December 2014
METHOD SUMMARY
1. Draw a diagram, and label it with variables.
2. Write a function for the quantity to be optimized, in terms of the variables you
have defined.
3. Use geometric formulae to express the function in terms of only one variable.
4. Take the derivative of your function.
5. Solve the derivative for zero.
6. Use the answer to determine the optimum quantity, and the variable values that
would produce it.
FARMER CHAN
Farmer Chan (Janey Sue’s husband) has 1,000 metres of fencing wire, and wants to
make the biggest field possible for his cows to graze in. There is a long, straight
existing fence that he can use for one side of the new field. What is the area of the
biggest field that he can make, and what should its side lengths be?
First, draw a diagram of the situation, and
label the variables.
Then, come up with an expression for the
quantity you are trying to maximize, in terms of
those variables.
L
W
𝐴 = 𝐿𝑊
FARMER CHAN
We would then want to maximize the area, A. In other words, we need to find the
derivative, A’, and find where A’ equals zero. This will give us the local maximum or
minimum (or possibly just an inflection point, but hopefully not!).
The problem is that we cannot take a derivative when there are two variables, L and
W. So, we need a geometric formula that lets us state one variable in terms of the
other. Try to work it out yourself first.
𝐿 + 𝑊 + 𝑊 = 1000 (because he has 1000 metres of wire total)
𝐿 + 2𝑊 = 1000
𝐿 = 1000 − 2𝑊
FARMER CHAN
𝐿 = 1000 − 2𝑊
Insert this into the original formula to express A in terms of one variable only.
𝐴 = 𝐿𝑊
𝐴 = 1000 − 2𝑊 𝑊
𝐴 = −2𝑊 2 + 1000𝑊
Find the derivative, A’, with respect to W as the independent variable.
𝐴′ = −4𝑊 + 1000
FARMER CHAN
𝐴′ = −4𝑊 + 1000
Solve the derivative for zero, because this will be a critical point.
0 = −4𝑊 + 1000
4𝑊 = 1000
𝑊 = 250
Therefore, when W is 250 metres, and L is 500 metres, the area A will be 125,000
square metres. This critical point is a local maximum… you can prove this by seeing
what happens if W was instead 249 or 251 metres. Problem solved!
TIN CAN
A tin can is being manufactured, and we want to use the least possible amount of
metal to hold a given volume. Let’s say we want a volume of 500 mL. What would
be the dimensions of a can that uses the least possible metal to hold that?
Draw a diagram and label it.
Q: What quantity is being optimized?
r
A: The surface area – in this case minimized.
𝑆 = 2𝜋𝑟 2 + 2𝜋𝑟ℎ
h
TIN CAN
𝑆 = 2𝜋𝑟 2 + 2𝜋𝑟ℎ
We have two variables in our formula for S, and we
need to work with just one.
Q: What geometric formula could link r and h? What
other information was given that could help?
r
A: The volume must be exactly 500 mL.
𝑉 = 𝜋𝑟 2 ℎ = 500
ℎ=
500
𝜋𝑟 2
h
TIN CAN
Substitute one expression in to the other.
𝑆 = 2𝜋𝑟 2 + 2𝜋𝑟ℎ
𝑆 = 2𝜋𝑟 2 + 2𝜋𝑟
𝑆 = 2𝜋𝑟 2 +
500
𝜋𝑟 2
1000
𝑟
r
Q: What is the next step?
A: Take the derivative of S. Try to do this on your
own before going to the next slide.
h
TIN CAN
𝑆 = 2𝜋𝑟 2 +
1000
𝑟
𝑆 ′ = 4𝜋𝑟 − 1000𝑟 −2
Now set the derivative to zero, and solve. Try to do
this on your own before advancing.
r
0 = 4𝜋𝑟 − 1000𝑟 −2
0 = 4𝜋𝑟 3 − 1000
1000 =
250
𝜋
=
(multiplied both by r2)
h
4𝜋𝑟 3
𝑟3
… therefore 𝑟 =
3
250
𝜋
≈ 4.301
TIN CAN
Q: By the way, if the volume is measured in mL, what is
the radius and height measured in?
A: Centimetres… 1 mL = 1 cm3
Radius is 4.301 cm. Also work out the height.
ℎ=
500
𝜋𝑟 2
500
=
𝜋
3 250
𝜋
2
≈ 8.603 cm
Interesting fact: a can that holds the most volume for the
least metal will always be exactly as high as its
diameter.
r
h
NETWORK CABLE
Betsy Ann Chan (Janey Sue’s indolent sister) wants to run a network cable out to the
fields so she can have better wireless access for her frequent breaks. The cable
needs to run from Point A to Point B; from one river bank to the other, and slightly
downstream. It costs $2 per metre to run the cable across land, and $5 per metre to
run it along the riverbed. The dimensions are given in the diagram.
10 metres
4 metres
A
B
NETWORK CABLE
NETWORK CABLE
Let’s show a general case for the cable, and try to define some variables. We’ll add
them on to the existing diagram.
What dimensions would you choose on the diagram for the variables?
10 m
x
4m
A
B
y
NETWORK CABLE
NETWORK CABLE
See if you can write a function for cable cost before you advance. Remember it is $5
to go through the water, and only $2 to go along the riverbank.
Did you actually try this? Just checking! Make sure you wrote a function for cable
cost and not just cable length.
𝐶 = 5 42 + 𝑥 2 + 2𝑦
10 m
x
4m
A
B
y
NETWORK CABLE
NETWORK CABLE
This geometric formula is actually pretty easy, and almost could have just been part
of the original diagram. There are two possibilities, and see if you can get it before
you click.
𝑥 = 10 − 𝑦
or
10 m
𝑦 = 10 − 𝑥
x
4m
A
B
y
NETWORK CABLE
It is probably easier to get rid of y, because the algebra will be simpler. Put it back
in the function C…
𝐶 = 5 16 + 𝑥 2 + 2𝑦
𝐶 = 5 16 + 𝑥 2 + 2 10 − 𝑥
𝐶 = 5 16 + 𝑥 2 + 20 − 2𝑥
10 m
x
4m
A
B
y
NETWORK CABLE
𝐶 = 5 16 + 𝑥 2 + 20 − 2𝑥
The fourth step is to take the derivative, C’. Do this yourself before you advance the
slides.
10 m
x
4m
A
B
y
NETWORK CABLE
𝐶 = 5 16 + 𝑥 2 + 20 − 2𝑥
′
𝐶 =5∙
′
1
2
1
16 + 𝑥
𝐶 = 5𝑥 16 + 𝑥
2 −2
∙ 2𝑥 − 2
1
2 −2
−2
Our fifth step is to then make C’
zero, and solve for x.
10 m
x
4m
A
B
y
NETWORK CABLE
1
0 = 5𝑥 16 + 𝑥
2=
2 −2
−2
5𝑥
16+𝑥 2
25𝑥 2
4=
16+𝑥 2
64 + 4𝑥 2 =
(square both sides – sneaky!)
25𝑥 2
10 m
B
64 = 21𝑥 2
64
21
= 𝑥2
∴𝑥=
x
4m
64
21
(discard negative)
A
y
NETWORK CABLE
Sixth and final step is to answer the question fully.
∴𝑥=
64
21
≈ 1.75 m
∴ 𝑦 ≈ 8.25 m
𝐶 = 5 16 + 𝑥 2 + 20 − 2𝑥
10 m
∴ 𝐶𝑚𝑖𝑛 = $38.33
Betsy Ann gets
x
4m
A
her wireless! y
B
CARDBOARD BOX
A six-sided cardboard box needs to have a volume of exactly 15 cubic feet. The top
and bottom of the box are to be exactly square. Work out what the dimensions of
the box must be for it to use the least amount of cardboard.
For this one you’re completely on your own!
All answers on the next slide.
CARDBOARD BOX
𝑆 = 4ℎ𝑠 + 2𝑠 2
(expression for surface area)
𝑉 = 15 = ℎ𝑠 2
(the geometric expression that links variables)
15
𝑠2
60
∴ 𝑆 = + 2𝑠 2
𝑠
60
𝑆 ′ = − 2 + 4𝑠
𝑠
∴ℎ=
𝑠=
3
15 ≈ 2.466 ft
s
(the surface area with one variable)
(derivative of surface area)
(solved for zero)
∴ ℎ ≈ 2.466 ft
𝑆 ≈ 36.493 ft2 (optimal surface area)
h
s
WRAPPING UP