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Appendix C C-11<br />

MINI PROJECT<br />

An Inequality for the Garden<br />

Part of a landscape architect’s plans call for a rectangular garden with a<br />

perimeter of 150 ft. The architect wants to find the width w and length l that<br />

maximize the area A of the garden. Two preliminary sketches are shown in the<br />

figure that follows. (In both cases, note that the perimeter is 150 ft, as required.)<br />

Of the two sketches, the one on the right yields the larger area. In this<br />

project you’ll use the inequality given in Exercise 40(b)* to find exactly which<br />

combination of w and l yields the largest possible area.<br />

(a) As background, first review or rework Exercise 40(b). If you are working<br />

in a group, assign one person to do this and then present the result to the<br />

others.<br />

25 ft<br />

30 ft<br />

50 ft<br />

45 ft<br />

A=25 ft 50 ft=1250 ft@ A=30 ft 45 ft=1350 ft@<br />

(b) Letting w denote the width of the rectangular garden, and assuming that<br />

the perimeter is 150 ft, show or explain why the area A is given by<br />

A w(75 w).<br />

(c) Use the inequality in Exercise 40(b) to show that A 56254 1406.25.<br />

In words: Given a perimeter of 150 ft, no matter how we pick w and l,<br />

the area can never exceed 1406.25 ft 2 . Hint: Apply the inequality to the<br />

product w(75 w).<br />

(d) From part (c) we know that the area can’t exceed 1406.25 ft 2 , so now<br />

the job is to find a combination of w and l yielding that area. To accomplish<br />

this, take the equation obtained in part (b), replace A with 1406.25,<br />

and then use a graphing utility to solve for w.<br />

(e) As a check on the work with the graphing utility (and to stay in shape with<br />

the algebra), use the quadratic formula to solve the same equation.<br />

(f) Now that you know w, find l and describe the dimensions of the required<br />

rectangle.<br />

*Exercise Set 2.3, Precalculus: A Problems-Oriented Approach, 7th edition

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