Quadratic "Max/Min" Word Problems (page 3 of 3)

Sections: Projectile motion, General word problems, Max/min problems

When you get to calculus, you will see some of these max/min exercises again. At that point, they'll want you to differentiate to find the maximums and minimums; at this point, you'll find the vertex, since the vertex will be the maximum or minimum of the related graphed parabola. But they're the same exercise and you'll get the same answers then as you will now.

• You have a 500-foot roll of fencing and a large field. You want to construct a rectangular playground area. What are the dimensions of the largest such yard? What is the largest area?

The fencing-length information gives me perimeter. If the length of the enclosed area is L and the width is w, then the perimeter is 2L + 2w = 500, so L = 250 – w. By solving the perimeter equation for one of the variables, I can substitute into the area formula and get an equation with only one variable:

A = Lw = (250 – w)w = 250w – w² = –w² + 250w

To find the maximum, I have to find the vertex (h, k).

h = ^–b/_2a = –(250)/2(–1) = –250/–2 = 125

In my area equation, I plug in "width" values and get out "area" values. So the h-value in the vertex is the maximizing width, and the k-value will be the maximal area:

k = A(125) = –(125)² + 250(125) = –15 625 + 31 250 = 15 625

The problem didn't ask me "what is the value of the variable w?", but "what are the dimensions?" I have w = 125. Then the length is L = 250 – w = 250 – 125 = 125.

The largest area will have dimensions of 125' by 125',
for a total area of 15 625 square feet.

Note that the largest rectangular area was a square. This is always true: for a given perimeter, the largest rectangular area will be that of a square. However, teachers are starting to notice that students have figured this out, so they're giving more complicated area-perimeter problems.

• You have a 1200-foot roll of fencing and a large field. You want to make two paddocks by splitting a rectangular enclosure in half. What are the dimensions of the largest such enclosure?
     

I'm dealing with something that looks like this:    It doesn't really matter which side I label as the "length" and which I label as the "width", as long as I label clearly and work consistently.

With the labelling I've chosen, the fencing gives me a "perimeter" of 2L + 3w = 1200. Solving for one of the variables, I get:

2L + 3w = 1200
L + 1.5w = 600
L = –1.5w + 600

Then the area is:   Copyright © Elizabeth Stapel 2004-2011 All Rights Reserved

A = Lw = (–1.5w + 600)w = –1.5w² + 600w

To maximize this area, I have to find the vertex. Since all I need are the dimensions (not the area), all I need from the vertex (h, k) is the value of h, since this will give me the maximal width.

h = ^–b/_2a = –(600)/2(–1.5) = –600/–3 = 200

Then the length will be L = –1.5(200) + 600 = –300 + 600 = 300.

The paddock should be 300' by 200', with the divider
running parallel to the 200-foot-long side.

Notice that, with the divider running down the middle of the paddock, you don't get a square as being the maximal shape. If they throw in cost considerations (like putting prettier but more expensive fencing on the side of the paddock facing the street), you'll get odd-sized results, too. Warning: Don't just assume that the maximal rectangular shape will always be a square.

In many quadratic max/min problems, you'll be given the formula you need to use. Don't try to figure out where they got it from. Just find the vertex. Then interpret the variables to figure out which number from the vertex you need, where, and with what units.

• Your factory produces lemon-scented widgets. You know that each unit is cheaper, the more you produce. But you also know that costs will eventually go up if you make too many widgets, due to the costs of storage of the overstock. The guy in accounting says that your cost for producing x thousands of units a day can be approximated by the formula C = 0.04x² – 8.504x + 25302. Find the daily production level that will minimize your costs.

As you might guess, it will be a lot easier to use the vertex formula to find the minimizing value for this quadratic than it would be to complete the square. So I'll use the vertex formula:

h = ^–b/_2a = –(–8.504)/2(0.04) = 8.504/0.08 = 106.3

Since the inputs to the formula they gave me are the production levels ("thousands of units"), I've found the number I need. The other number for the vertex would be the actual costs for making this amount of widgets, and the problem doesn't ask for that. I do need to remember, though, that x is in thousands of units, so my best level of production is not 106.3 units, but (106.3)(1 000) = 106 300 units.

I will minimize my costs if I produce 106 300 units a day.

Sometimes you'll get hit with a problem that seems much more complicated, especially when you have to invent the formula yourself.

• You run a canoe-rental business on a small river in Ohio. You currently charge $12 per canoe and average 36 rentals a day. An industry journal says that, for every fifty-cent increase in rental price, the average business can expect to lose two rentals a day. Use this information to attempt to maximize your income. What should you charge?

Let's say I have no idea how to set this problem up. Instead of going straight to an equation, I'll need to put in some real numbers, see what I do when I know what the values are, and then follow the pattern to get my formula. Here is my reasoning, neatly laid out in a table:

price hikes	price per rental	number of rentals	total income / revenue
none	$12.00	36	$12.00×36 = $432.00
1 price hike	$12.00 + 1(0.50)	36 – 1(2)	$12.50×34 = $425.00
2 price hikes	$12.00 + 2(0.50)	36 – 2(2)	$13.00×32 = $416.00
3 price hikes	$12.00 + 3(0.50)	36 – 3(2)	$13.50×30 = $405.00
x price hikes	$12.00 + x(0.50)	36 – x(2)	(12 + 0.5x)(36 – 2x)

Then my formula for my revenues R after x fifty-cent price hikes is:

R(x) = (12 + 0.5x)(36 – 2x) = 432 – 6x – x² = –x² – 6x + 432

The maximum income will occur at the vertex of this quadratic's parabola, and the vertex is at
(–3, 441):

h = ^–b/_2a = –(–6)/2(–1) = 6/(–2) = –3
k = R(h) = –(–3)2 – 6(–3) + 432 = –9 + 18 + 432 = 450 – 9 = 441

That is, my income will be maximized (assuming the journal article is correct) if I lower my current price of $12 by three times of fifty cents, or by $1.50.

I should charge $10.50 per canoe.

Whenever you're not sure of your formula, try doing what I did above: write out what you would do if you knew what the numbers were, and see if you can turn that into a formula. But make sure you write things out completely, like I did, so you can see the pattern. Warning: Don't simplify too much in your head, or you could miss what your formula is supposed to be.

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