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Optimization
Optimization is an application of derivatives that deals with finding the minimum or maximum
values using some kind of constraint equation. More often than not, these will be word problems.
Steps to optimization:
1.
2.
3.
4.
5.
Identify the constraint.
Identify the variable that is to be maximized or minimized.
Form equations based on steps one and two.
Solve the equations using various methods.
Answer the initial question.
Let’s run through an example problem.
Example 1: Farmer Jaime has 1,000 feet of fencing to contain his children as he teaches them kalkoolus.
He wants to maximize the amount of area that he can enclose. One side of the area is up against a river
and does not require fencing. What dimensions of the fence will maximize the area enclosed?
Drawing a diagram is an effective way to understand the problem.
River
y
x
However, a simple diagram will do fine. For this example, the Area is being maximized and the amount
of fencing is the constraint. Thus, we can arrive at the following equations:
A = xy (Maximize)
1000 = x + 2y (Constraint)
The first equation is the standard area equation for any rectangle. The second equation is a modified
form of the perimeter equation. There is only one side of the length x thanks to the river, whereas there
are two sides of the length y. Now, notice that there are at least two unknown variables in each
equation. This means that you will need to substitute one equation into the other. With this in mind,
solve the constraint equation for x.
x = 1000 – 2y
Next, substitute this new equation into the area equation. This eliminates the variable x and allows you
to solve for y.
A = (1000 – 2y)y
A = 1000y – 2y²
In order to find any form of extrema, you must take a derivative and set it equal to zero. Taking the
derivative of the area equation, you get the following:
A’ = 1000 – 4y = 0
Solving for the critical point(s):
0 = 1000 – 4y
4y = 1000
y = 250
Now that we have one of the variables, we can plug it into the constraint equation and figure out the
other one:
1000 = x + 2y
1000 = x + 2(250)
1000 = x + 500
x = 500
Now, be sure the question has been answered.
The dimensions of the fenced-in area are 500 feet by 250 feet.
Congratulations! You have completed an optimization problem. Don’t forget, the solutions of
optimization problems tend to be absolute maximums or absolute minimums of a function.
Let’s see another common example.
Example 2: The kids in Jaime’s kalkoolus class wish to construct a box to hold a gift for him. Due to the
nature of the gift, the box’s length must be twice as long as its width. Material to wrap the box costs $5
per square foot (assume they are wrapping all 6 sides and not using any extra material). If the volume of
the box must be 10 cubic feet, determine the minimum cost of wrapping the box.
Well, that’s certainly a doozy of a problem. As always, a safe way to start is to draw a diagram. Labels
are helpful here.
Height = h
Width = w
Length = 2w
Fantastic. That is a quality diagram. Good work, young scholar. Now, create the constraint equation and
maximum/minimum equation. In our case here, we are trying to minimize the cost. Let’s start with the
easy equation, the constraint. The typical volume formula for a rectangular prism is:
V = lwh
Substituting what we know and simplifying:
10 = (2w)wh
10 = 2w²h (constraint)
Good. Now, let’s tackle the second equation. We are trying to minimize the cost. The cost is dependent
on surface area. Let’s look at the surface area equation for a rectangular prism:
SA = 2lw + 2lh + 2hw
The cost is $5 per square foot. Using this, we can create a “cost” equation:
C = 5(SA)
C = 5(2lw + 2lh + 2hw)
Factoring out a 2 to make the equation simpler, we arrive at:
C = 10(lw + lh + hw)
If we substitute what we know into the cost equation…
C = 10(2w² + 2wh + hw) (cost)
So, here are our equations:
C = 10(2w² + 2wh + hw) (cost)
10 = 2w²h (constraint)
In order to determine the minimum of a function, we need to take a derivative. We are minimizing the
cost, so we will take the derivative of that equation. Because we are high schoolers and hate work, we
do not wish to use the product rule. Besides, it would be difficult to find a critical value with two
unknowns. To rectify this issue, we can solve the constraint equation for h and plug its value into the
cost equation.
h
10
(pardon the different font for these fractions)
2w 2
C  10(2w 2  2w(
10
10
)  ( 2 ) w)
2
w
w
Simplifying…
C  20 w 2 
C  20w 2 
200 100

w
w
300
or C  20w 2  300w 1
w
Looking good. Now, take a derivative:
C   40w  300w 2 or C   40w 
300
w2
Setting the first derivative equal to zero, we can solve for critical points:
300
w2
40w 3  300
0
w2
0  40w 
Zero is a critical value because it would result in an undefined value. However, the box cannot have a
width of zero because it wouldn’t be able to hold anything! Keep solving…
0  40 w 3  300
300  40 w 3
300
 w3
40
15
w3
2
w = 1.957433821…
Victory! Now, how do we verify that this is a minimum? Well, you could just take my word for it, you
could be a teenager and not care, you could test the changing of signs of the first derivative, or you
could use the sign of the second derivative to figure it out. Because the latter two are extremely lengthy,
we will not go through them here. Just be aware that the first derivative will go from negative to
positive and the second derivative will be positive to verify a minimum.
Now that we have w, we can easily find out l and h.
l = 2w
l = 2(1.957433821)
l = 3.914867641
h
10
2w 2
10
2(1.957433821) 2
h  1.30495588
h
Huzzah! With all three dimensions known, we can simply plug into any form of the cost equation to
answer the problem. Let’s use the original form to minimize mistakes.
C = 10(lw + lh + hw)
C = 10((3.914867641)( 1.957433821) + (3.914867641)(1.30495588) + (1.30495588)(1.957433821))
C = 10(15.32618865)
C = 153.2618865
C ≈$153.26
That’s pretty expensive, but at least you know your optimization!
Example 3: Find the x-value of the point on the graph f  x  
x that is closest to the point (3,0).
For this problem, we are minimizing distance. Remember the distance formula:
d
x2  x1 2   y2  y1 2
Since we are dealing with the problem (3,0), plug this into the formula.
d
 x 2  3 2   y 2  0  2
d
x  32  y 2
We need to eliminate y, right? Well, since f(x)is another way of saying y, we can replace y with the
original equation, getting…
d
d
x  32  x
x  32  x 2
2
Now, logically, if you minimize what is under the radical, you minimize the square root of it. To make the
rest of the problem easier, ignore the radical.
d  ( x  3) 2  x 2
Factoring this, we get:
d  2x 2  6x  9
To find the minimum of this equation, we need to take the derivative.
d   4x  6
From here, it should be self explanatory.
0  4x  6
4x  6
x3
2
This means that at the x-value of x = 3/2 on the graph of x , the shortest distance between x and
(3,0) occurs. If the problem asked for it, you could also find the y-value by plugging it back into the
original equation.
Some problems will be extremely easy, while others will be brutally hard. The only way to get good at
optimization problems is to familiarize yourself with the different types of problems. On that note, try
the following:
Problem 1: The previous box was too expensive for Jaime’s kalkoolus class. Instead, they are going to
use a 20 square foot piece of colorful cardboard to make a box. If all the material is to be used and the
box is to have a square base and no top, what is the maximum volume the box can have? Is the box
going to be large enough to hold the gift? (Note: The previous box needed to be 10 cubic feet.)
Diagram:
Problem 2: Jaime is trying to recruit more kids for his kalkoolus class. In order to do so, he is printing a
rather large poster. His sheet of paper is 100 square feet. The printer has 5 inch margins on the side, a 4
inch margin on the top, and a 6 inch margin on the bottom. What is the maximum area of the printed
part of the poster?
Problem 3: The sum of two nonnegative numbers is 21. Find the value of these numbers so that the
product of one of the numbers times three times the other number is a maximum.
Problem 4: Find the point (x,y) on the graph of x³ that is closest to the point (7,0).
(Hint: use the distance formula.)