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OPTIMATIZATION - MAXIMUM/MINIMUM PROBLEMS – BC CALCULUS
1. Read each problem slowly and carefully. Read the problem at least three times before trying to solve it.
Sometimes words can be ambiguous. It is imperative to know exactly what the problem is asking. If you
misread the problem or hurry through it, you have NO chance of solving it correctly.
2. If appropriate, draw a sketch or diagram of the problem to be solved. Pictures are a great help in
organizing and sorting out your thoughts.
3. Define variables to be used and carefully label your picture or diagram with these variables. This step is
very important because it leads directly or indirectly to the creation of mathematical equations.
4. Write down all equations which are related to your problem or diagram. Experience will show you that
MOST optimization problems will begin with two equations. One equation is a "constraint" equation and
the other is the "optimization" equation – the one you are asked to maximize or minimize. The
"constraint" equation is used to solve for one of the variables. This is then substituted into the
"optimization" equation before differentiation occurs. Some problems may have NO constraint equation.
Some problems may have two or more constraint equations.
5. Before differentiating, make sure that the optimization equation is a function of only one variable. Then
differentiate using the well-known rules of differentiation.
Example 1:
Two positive numbers have a sum of 60. What is the maximum product of one number times the
square of the second number?
Constraint eq.
Optimization eq.
𝑎 + 𝑏 = 60
𝑃 = 𝑎2 𝑏
𝑃𝑚𝑎𝑥 =?
𝑃 = 𝑎2 𝑏 = 𝑎2 (60 − 𝑎) = 60𝑎2 − 𝑎3
𝑑𝑃
= 120𝑎 − 3𝑎2 = 0 ⇒ 𝑎(40 − 𝑎) = 0
𝑑𝑎
𝑎 = 0 𝑃 = 0 𝑚𝑖𝑛𝑖𝑚𝑢𝑚
𝑎 = 40 𝑏 = 20 𝑃𝑚𝑎𝑥 = 32000
Example 2:
We need to enclose a portion of a field with a rectangular fence. We have 500 feet of fencing
material and a building is on one of the longer sides of the field and so won’t need any fencing.
Determine the dimensions of the field that will enclose the largest area.
Constraint eq.
𝑎 + 2𝑏 = 500
Optimization eq.
𝐴 = 𝑎𝑏
𝐴𝑚𝑎𝑥 =?
𝐴 = 𝑎𝑏 = (500 − 2𝑏)𝑏 = 500𝑏 − 2𝑏 2
𝑑𝐴
= 500 − 4𝑏 = 0 ⇒ 𝑏 = 125 𝑎 = 250
𝑑𝑎
𝐴𝑚𝑎𝑥 = 31250
2
Example 3:
We need to enclose a portion of a field with a rectangular fence that has two pens of equal area and
equal size. We want to enclose at total of 10,000 square feet. Determine the least amount of fencing
we will need.
Constraint eq.
Optimization eq.
𝐴 = 2𝑎𝑏 = 10000
𝑃 = 3𝑏 + 4𝑎
𝑃𝑚𝑖𝑛 =?
𝑃 = 3𝑏 + 4𝑎 = 15000/𝑎 + 4𝑎
𝑑𝑃
15000
=−
+ 4 = 0 ⇒ 𝑎 = 61.237 𝑏 = 81.650
𝑎2
𝑑𝑎
𝑃𝑚𝑖𝑛 = 489.884 𝑓𝑒𝑒𝑡
Example 4:
We need to enclose a portion of a field with a rectangular fence. A river is on one side of the
fence, so we won’t need to fence along the river. We would like to enclose 10,000 square feet.
Because of the added construction costs of building perpendicular to the river, it costs $8 a liner
foot to buy and build the fence. The portion of the fence parallel to the river only cost $5 per liner
foot to buy and build. What is the least amount of money we will need to construct our fence?
Constraint eq.
𝑎𝑏 = 10000
Optimization eq.
𝑀 = 2 × 8𝑏 + 5𝑎
𝑀𝑚𝑖𝑛 =?
𝑀 = 16𝑏 + 5𝑎 = 16𝑏 + 50000/𝑏
50000
𝑑𝑀
= 16 −
= 0 ⇒ 𝑏 = 55.902 𝑎 = 178.885
𝑑𝑏
𝑏2
𝑀𝑚𝑖𝑛 = $1789.147
Example 5:
We want to construct a box with a square base and we only have 10 square meters of material to use
in construction of the box. Assuming that all the material is used in the construction process
determine the maximum volume that the box can have.
Constraint eq.
2𝑎2 + 4𝑎𝑏 = 10
Optimization eq.
𝑉 = 𝑎2 𝑏
𝑉𝑚𝑎𝑥 =?
5 𝑎
5𝑎
𝑉 = 𝑎2 𝑏 = 𝑎2 � − � =
− 𝑎3 /2
2𝑎 2
2
𝑑𝑉 5 3𝑎2
5
= −
=0 ⇒𝑎=�
𝑑𝑎 2
2
3
𝑉𝑚𝑎𝑥 = 𝑎3 =
5 5
�
3 3
5
𝑏=�
3
3
Example 6:
A cylindrical aluminum soda can is to hold 12 ounces of tasty beverage. What are the dimensions of
the can that will minimize the amount of aluminum used? Hint:
12 US fluid ounces equals 21.6562 cubic inches. If the dimensions of the actual can are 2.6 inch
diameter and 4.75 inches tall, does the actual can have the optimal dimensions?
Constraint eq.
Optimization eq.
(𝜋𝑟 2 )ℎ = 21.6562
𝐴 = (2𝜋𝑟)ℎ + 2(𝜋𝑟 2 )
𝐴𝑚𝑖𝑛 =?
𝐴 = (2𝜋𝑟)ℎ + 2(𝜋𝑟 2 ) = 2𝜋𝑟
43.3124
𝑑𝐴
=−
+ 4𝜋𝑟
𝑟2
𝑑𝑎
STRANGE ???
21.6562
43.3124
+ 2𝜋𝑟 2 =
+ 2𝜋𝑟 2
2
𝜋𝑟
𝑟
𝑟 = 1.511" ℎ = 15.664"
𝐴𝑚𝑖𝑛 = 163.057 𝑖𝑛𝑐ℎ𝑒𝑠 3
Sometimes we also want to maximize CLOSENESS!
Example 7:
Determine the point(s) on 𝑦 = 𝑥 2 + 1 that are closest to (0,2).
𝑃𝑜𝑖𝑛𝑡𝑠 (𝑥, 𝑦)𝑎𝑛𝑑 (0,2)
Constraint eq.
𝑦 = 𝑥2 + 1
Optimization eq.
𝑑 = �(𝑥 − 0)2 + (𝑥 2 + 1 − 2)2
𝑑 = �𝑥 4 − 𝑥 2 + 1
𝑑𝑑
4𝑥 3 − 2𝑥
=
=0
𝑑𝑥
2√𝑥 4 − 𝑥 2 + 1
𝑓𝑜𝑟 𝑥 = 0, 𝑑 = 1
⇒ 𝑑𝑚𝑖𝑛 =
√3
2
𝑑𝑚𝑖𝑛 =?
⇒ 2𝑥(𝑥 2 − 1)
𝑓𝑜𝑟 𝑥 = ±
√2
2
,𝑑 =
√3
2
𝑥=0
√2 3
√2 3
𝑓𝑜𝑟 � , � 𝑎𝑛𝑑 �−
, �
2 2
2 2
𝑥=±
√2
2
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At this point, you might be tempted to think that the optimal value is always at the
critical value. HA HA HA
Example 8:
A 2 feet piece of wire is cut into two pieces and once piece is bent into a square and the other is bent
into an equilateral triangle. Where should the wire cut so that the total area enclosed by both is a (a)
minimum and (b) maximum? (Note: it is not required to build both shapes).
Let’s cut the wire into two pieces. The first piece will have length x which we’ll bend into a square and each
𝑥
4
side will have length . The second piece will then have length (2 − 𝑥)(constraint) and we’ll bend this into
an equilateral triangle and each side will have length
1
(2 −
3
𝑥).
Optimization eq.
𝑥 2 1 1
𝑥 2 √3
√3
𝐴(𝑥) = � � + � (2 − 𝑥)� � (2 − 𝑥)� =
+
(2 − 𝑥)2
6
16 36
4
2 3
𝐴𝑚𝑎𝑥 =?
𝑑𝐴 𝑥 √3
8√3
𝑥 √3
√3
(2)(2 − 𝑥)(−1) = +
= +
𝑥−
=0 ⇒ 𝑥=
= 0.8699
9
𝑑𝑥 8 36
8 18
9 + 4√3
𝐴𝑚𝑖𝑛 =?
Now, let’s notice that the problem statement asked for both the minimum and maximum enclosed
area and we got a single critical point. BUT:
Let’s notice that x must be in the range 0 ≤ 𝑥 ≤ 2 and since the area function is continuous we use
the basic process for finding absolute extrema of a function.
𝐴(0) = 0.1925
𝐴(0.8699) = 0.1087
𝐴(2) = 0.25
So, it looks like the minimum area will arise if we take 𝑥 = 0.8699 while the maximum area will
arise if we take the whole piece of wire and bend it into a square.
One of the extrema in the previous problem was at an endpoint and that will happen on occasion.
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Sometimes we need to establish a pattern to write our primary equation (Examples 8 & 9).
Example 9:
The manager of an 80-unit apartment complex is trying to decide what rent to charge. Experience
has shown that at a rent of $200, all of the units will be full. On the average, one additional unit will
remain vacant for each $20 increase in rent. Find the rent to charge so as to maximize revenue.
a pattern:
a) 𝑥 = the number of $20 increases. The rent for an apartment is 200 + 20𝑥
b) number of apartments rented:
For each $20 increase, one less apartment will be rented. Since 𝑥 is the number of 20 increases,
the number of vacancies will be 𝑥, and the number of apartments rented will be
80 − 𝑥
Optimization eq.
the total revenue from all rented apartments:
𝑅 = (80 − 𝑥)(200 + 20𝑥) = 16000 + 1400𝑥 − 20𝑥 2
𝑑𝑅
= 1400 − 40𝑥 = 0
𝑑𝑥
⇒
𝑥 = 35
𝑅𝑚𝑎𝑥 =?
Rent for an apartment would be $900 (𝑣𝑒𝑟𝑦 𝑒𝑥𝑝𝑒𝑛𝑠𝑖𝑣𝑒), 45 𝑟𝑒𝑛𝑡𝑒𝑑 𝑎𝑛𝑑 𝑅 = $40500
Example 10:
A fence, 8 feet high, is parallel to the wall of a building and 1 foot
from the building. What is the shortest plank that can go over the
fence, from the level ground, to prop the wall?
Constraint eq.
sin 𝜃 =
8
𝑥
1
𝑦
8
1
=
+
sin 𝜃 cos 𝜃
cos 𝜃 =
Optimization eq.
𝐿 =𝑥+𝑦
0<𝜃<
𝜋
2
𝑑𝐿 −8 cos 𝜃 sin 𝜃
=
+
= 0 ⇒ −8𝑐𝑜𝑠 3 𝜃 + 𝑠𝑖𝑛3 𝜃 = 0
𝑑𝜃
𝑠𝑖𝑛2 𝜃
𝑐𝑜𝑠 2 𝜃
𝑡𝑎𝑛3 𝜃 = 8 ⇒ 𝑡𝑎𝑛 𝜃 = 2 ⇒ 𝜃 = 63.4𝑜 ≈ 1.1
𝑑𝐿
𝑑𝜃
𝜃
0
−
1.1
+
𝜋
2
6
𝜋
�0, � 𝑖𝑠 𝑜𝑝𝑒𝑛 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙, 𝑠𝑜 𝐿 ℎ𝑎𝑠 𝑙𝑜𝑐𝑎𝑙 𝑎𝑛𝑑 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑜𝑛𝑙𝑦 𝑎𝑡 𝜃 = 1.1
2
𝑥 = �82 + �
8 2
1
� = 8�1 + = 4√5
tan 𝜃
4
𝑦 = �12 + (1 × tan 𝜃)2 = √1 + 4 = √5
𝐿 = 𝑥 + 𝑦 = 5√5𝑓𝑡 ≈ 11.2𝑓𝑡
It’s fun to inscribe shapes into other shapes that circumscribe the inscribed shape we’re inscribing.
Example 11:
Find the volume of the largest cone that can be inscribed inside a sphere of
radius 5.
Constraint eq.
Optimization eq.
(ℎ − 𝑅)2 + 𝑟2 = 𝑅2
𝑉 = 1/3 𝜋 𝑟2 ℎ
1
1
𝑉 = 𝜋 [𝑅 2 − (ℎ − 𝑅)2 ]ℎ = 𝜋 [2𝑅ℎ2 − ℎ3 ]
3
3
𝑑𝑉 1
= 𝜋[4𝑅ℎ − 3ℎ2 ] = 0
𝑑ℎ 3
32
𝑉𝑚𝑎𝑥 = 81 𝜋𝑅 3
⇒ ℎ = 0 (𝑚𝑖𝑛)
ℎ=
4
𝑅 (𝑚𝑎𝑥)
3
Example 12:
An arch top window is being built whose bottom is a rectangle and the top is a semicircle.
If there is 12 meters of framing materials what is the width of the window that lets in
the most light?
Constraint eq.
2ℎ + 2𝑟 + 𝜋𝑟 = 12
Optimization eq.
1
𝐴 = 2ℎ𝑟 + 𝜋𝑟 2
2
1
1
1
𝐴(𝑟) = 2𝑟 �6 − 𝑟 − 𝜋𝑟� + 𝜋𝑟 2 = 12𝑟 − 2𝑟 2 − 𝜋𝑟 2
2
2
2
𝑑𝐴
= 12 − 𝑟(4 + 𝜋) = 0
𝑑𝑟
𝑑2𝐴
= −4 − 𝜋 < 0
𝑑𝑟 2
⇒
𝑟=
12
= 1.6803
4+𝜋
𝐴𝑚𝑎𝑥 𝑚𝑢𝑠𝑡 ℎ𝑎𝑣𝑒 𝑠𝑒𝑚𝑖𝑐𝑖𝑟𝑐𝑙𝑒 𝑟 = 1.6803 𝑎𝑛𝑑 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒 3.3606 × 1.6803(ℎ × 2𝑟)