Geometry Word Problems:
    Maximizing and Minimizing (page 6 of 6)

Sections: Introduction, Basic examples, Triangle formulas, Complex examples, The Box Problem & the Goat Problem, Max / Min problems

A special case of quadratic-based geometry word problems involves having 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.

• 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:

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.

There are a couple different ways of finding the vertex. I'll take the easy way. The equation of the quadratic, in y = ax² + bx + c 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

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 the "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:     Copyright © Elizabeth Stapel 2000-2011 All Rights Reserved

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.

Educators have started noticing that students have figured out the solution to the above exercise, just as a rule: "The rectangle with the largest area for a given perimeter will be a square" and, vice versa, "The rectangle with the shortest perimeter for a given area will be a square". So they've come up with new forms of the exercise; fortunately, the reasoning and general process is exactly the same.

• The riding stables has just received an unexpected rush of registrations for the next horse show, and quickly needs to create some additional paddock space. There is sufficient funding to rent 1200 feet of temporary chain-link fencing. The plan is to form two paddocks with one shared fence running down the middle. What is the maximum area that the stables can obtain, and what are the dimensions of each of the two paddocks?

To help me "see" what I'm doing, I first draw a picture:

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The total area A for the two paddocks will obviously be A = Lw. The total length of fencing gives me the total "perimeter" (in quotes, because I'm also including that shared line down the middle, so this isn't the "regular" perimeter). Then:

P = 1200 = 2L + 3w

I can't do much with this, but what if I solve this for one of the variables, and then plug that into the "area" formula? Let's see...

1200 = 2L + 3w
1200 – 3w = 2L + 3w – 3w
1200 – 3w = 2L
(1200 – 3w) / 2 = (2L) / 2
600 – (3/2)w = L

Now I'll plug this into the "area" formula:

A = Lw = [600 – (3/2)w](w) = 600w – (3/2)w²
   = (–3/2)w² + 600w

This is a negative quadratic, just like the previous exercise, and I'll find the maximum area in the same way: by finding the vertex.

h = –b / (2a) = –(600) / [2(–3/2)] = –600 / –3 = 200

So I get the maximum area when the input (the width, in this case) has a value of 200. Reviewing my picture and equations, I see that the width should be 200 feet, the overall length should be 300 feet, and each paddock should then be 150 feet long.

The maximum added area will be 60,000 square feet (sq ft), 30,000 sq ft for each paddock. Each paddock should measure 200 ft by 150 ft, and the paddocks should share a 200-ft long side.

In particular, note that the maximal area above is not a square! Other ways of skewing the solutions away from squares, circles, or spheres is to include cost considerations, such as the material for the base of an open-topped box costing more (because it needs to be stronger) than the material for the sides. Don't just assume that the "neatest" answer will be correct. Do the math.

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