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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