Print-friendly page

Geometry Word Problems:
     Maximizing Area and Volume (page 4 of 4)

Sections: Introduction, Basic shapes, The Pythagorean Theorem, Max/min problems

• You need to make a pizza box. You know that the box needs to be two inches deep, it needs to be a square, and the web site you found said that the box needs to have a volume of 512 cubic inches. After cursing the occasional near-uselessness of the information you find on the Internet, you start calculating the dimensions you will need. You have a large piece of cardboard, but you don't have enough cardboard to make a mistake and try again, so you'll have to get it right the first time. You will be forming the box by cutting out a large square, and then cutting out the two-inch squares from the corners that will allow you to fold up the edges to make a two-inch-deep box. What should be the dimensions of the large square? (Ignore the top of the box:  you'll just make another open box, slightly larger, turn it upside down, and slip it over the first box to make the "top".)

Visually, this is what I'm doing:

	Your initial piece of cardboard. Its dimensions are x inches by x inches, but you don't know the value of x yet.
	The corners are cut out, leaving two-inch flaps on all four sides.
	The flaps are folded up to make a two-inch deep box.

Note that we will be letting "x" stand for the length and width of the original square of cardboard in the computations that follow. The value of x is what you're looking for.

Looking at the picture, you can see that the width of the bottom of this box will be x – 2 – 2; that is, the width will be the original x inches, minus two inches on either side because of the flaps that you're folding up to be the sides of the box. Then the width, simplified, is x – 4. The length is also x – 4. Since the depth of the box is 2 (that's the whole point of the two-inch-by-two-inch squares being cut out of the corners), you can then write the formula for the volume of this box:

volume = (length)(width)(depth) = 512 cubic inches
(x – 4)(x – 4)(2) = 512
2(x² – 8x + 16) = 512
x² – 8x + 16 = 256
x² – 8x – 240 = 0
(x – 20)(x + 12) = 0
x = 20  or  x = –12

I can ignore the extraneous negative result. Then x = 20 inches, and the answer is:

The large square should be twenty inches square.

There is a special case of quadratic area problem, where you have to maximize or minimize some dimension. This will involve finding the vertex of the quadratic formula you come up with, since the vertex will be the maximum or minimum of the graph. Here's an example:

• Find the largest possible rectangular area you can enclose, assuming you have 128 meters of fencing. What is the (geometric) significance of the dimensions of this largest possible enclosure?

I'll let the length be L and the width be W. I have 128 meters of fencing, so the perimter equation is: 

2L + 2W = 128

Dividing by 2 to make things simpler, I get:

L + W = 64

Previously, they would have given me the area and I would have had to find the length and width. This time, they told to find the area; in particular, to find the largest area, given this perimeter. How do I do that? Let's look at the area equation:

A = L × W

I can substitute for either one of these variables by solving the perimeter equation:

L + W = 64
L = 64 – W   (solving for L)

Then:   Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

A = (64 – W) × W   (substituting into the area equation)
     = 64W – W ²

In other words, my area equation is a quadratic, and I'm supposed to find the maximum. So all I really need to do is find the vertex. Since the above area equation is a negative quadratic, then it graphs as an upside-down parabola, so the vertex is the maximum.

(Max/min quadratic problems almost always work out this way: if you need to find the max, you'll get a negative quadratic; if you need to find the min, you'll get a positive quadratic. In either case, you solve the problem by finding the vertex.)

There are a couple different ways of finding the vertex. I'll take the easy way. The equation of the quadratic, in regular format, is:

A = –W ² + 64W

The vertex of a parabola is the point (h, k), where h = ^–b/_2a .  In this case:

h = ^–(64)/_(2×(–1)) = 32

To find the "k" part of the vertex, all I do is plug 32 in for W:

k = –(32)² + 64(32) = 1024

Remember that my points from this equation are (W, A) — that is, I plug in a width and figure out the area — so the "h" is the maximizing width and this "k" is the maximum area. So the answer is:

The largest possible area is 1024 square meters

...and I know that this maximum occurs when the width is 32 meters. Now I also need to find the length, because the original question asked about the significance of the dimensions. Since W = 32, then:

L = 64 – W = 64 – 32 = 32

Then the length and width are the same: 32 meters. What do you call a rectangle that is as wide as it is long? A square. So the second part of the answer is:

The largest possible rectangular area is in the shape of a square.

<< Previous  Top  |  1 | 2 | 3 | 4  |  Return to Index

  Copyright © 2006-2008  Elizabeth Stapel   |   About   |   Terms of Use

 Feedback   |   Error?