The Exciting World of
Optimization
Emma Sullivan
Optimization

Optimization is the procedure or procedures
used to make a system as effective or
functional as possible.

In this case, we will be using optimization to
maximize area and volume.
A rectangular patch of land is to be enclosed by
60 ft. of fencing for a doggie play pen. Find the
length and width that will give the maximum area.




The first step is choose
which formula fits the
problem.
Drawing a picture is
helpful.
For this example, the
formula for the
perimeter of a rectangle
is needed
A = 2L + 2w
A rectangular patch of land is to be enclosed by
60 ft. of fencing for a doggie play pen. Find the
length and width that will give the maximum area.



Next, plug in the information that the problem
gives you.
You know that the maximum fencing is 60, so
60 = 2L + 2w
Then solve for one of the variables.
2L = 60 – 2w
L = 30 – w
A rectangular patch of land is to be enclosed by
60 ft. of fencing for a doggie play pen. Find the
length and width that will give the maximum area.
Next, plug your L value back into the original
perimeter formula.
2(30 – w) + 2w = 60

Now since we are maximizing area, we need
to consider the area formula.
A= wL

A rectangular patch of land is to be enclosed by
60 ft. of fencing for a doggie play pen. Find the
length and width that will give the maximum area.
Using your values from 2(30 – w) + 2w = 60
Plug them into the area formula
A = 2(30 – w)(2w)

Simplify
A = (60 – 2w)(2w)
A = 120w – 4w2

A rectangular patch of land is to be enclosed by
60 ft. of fencing for a doggie play pen. Find the
length and width that will give the maximum area.
Now, take the derivative of the equation
A’ = 120 – 8w
 Set equal to zero and solve for w
0 = 120 – 8w
8w = 120
w = 15

A rectangular patch of land is to be enclosed by
60 ft. of fencing for a doggie play pen. Find the
length and width that will give the maximum area.
Finally, plug the value of w back into the original
perimeter equation to find the value of L
2L + 2(15) = 60
2L + 30 = 60
2L = 30
L = 15
 The length and width that will give the maximum
area are 15 ft X 15 ft.

An origami box is to be made by cutting squares from the
corners of a 20 x 20 cm square piece of paper and bending up
the sides. How long should the side of the square be so that
the box has a maximum volume?

Let each side of the soon-to-be cut out
squares be represented by x
Write the length, width, and height of the
paper in terms of x
w = 20 – x
L = 20 – x
H=x

An origami box is to be made by cutting squares from the
corners of a 20 x 20 cm square piece of paper and bending up
the sides. How long should the side of the square be so that
the box has a maximum volume?
Plug your new length, width, and height into
the formula for volume
V = Lwh
V = (20 – x)2 x
 Simplify
V = (4x2 – 80x + 400)x
V = 4x – 80x2 + 400x

An origami box is to be made by cutting squares from the
corners of a 20 x 20 cm square piece of paper and bending up
the sides. How long should the side of the square be so that
the box has a maximum volume?
Next, take the derivative
V’ = 12x2 – 160x + 400

Simplify
4(3x2 – 30x + 100)
4(3x – 10)(x – 10)

An origami box is to be made by cutting squares from the
corners of a 20 x 20 cm square piece of paper and bending up
the sides. How long should the side of the square be so that
the box has a maximum volume?
Set derivative equal to zero and solve for x
x = 10/3 and x = 10



Since the sides of the paper are only 20 cm,
the squares cannot be 10
Therefore, the answer is 10/3 cm.
You have now learned
optimization!
Give yourself a big pat
on the back!