College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 4 - Review Exercises - Page 317: 24

Answer

The largest area that can be enclosed is approximately 4,166,666.67 squared meters.

Work Step by Step

The perimeter of a rectangle is the addition of all its sides; however, there is an extra length, so: $P=2w+3l$ The area of a rectangle is width times length: $A=w\cdot l$ We want the area as a function of $l$, so we must re-write $w$ in terms of $l$. Since we know the value of the perimeter, we can solve for $w$: $10,000=2w+3l$ $10,000-3l=2w+3l-3l$ $10,000-3l=2w$ $(10,000-3l)/2=2w/2$ $w=5000-1.5l$ Now we can substitute $w$ and get the area as a function of its length: $A(l)=l\cdot (5000-1.5l)$ $A(l)=5000l-1.5l^2$ Now we'll re-write the quadratic equation to vertex form since the vertex is the global maximum: $A(l)=-1.5(l^2-\frac{10,000}{3}l)$ $A(l)=-1.5(l^2-\frac{10,000}{3}l+(\frac{\frac{10,000}{3}}{2})^2)+(\frac{\frac{10,000}{3}}{2})^2\cdot(1.5)$ $A(l)=-1.5(l^2-\frac{10,000}{3}l+(\frac{10,000}{6})^2)+\frac{10,000^2}{36}\cdot1.5$ $A(l)=-1.5(l-\frac{10,000}{6})^2+\frac{10,000^2}{24}$ So, the maximum area is the y-value of the vertex, which is about 4,166,666.67 squared meters.
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