College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 3 - Polynomial and Rational Functions - Exercise Set 3.1 - Page 345: 70


Dimensions of the playground: $100$ ft $\times 66.67$ ft Maximum area of playground: $6,666.67$ sq ft

Work Step by Step

*Refer back to page 340 for a complete reference on calculating maximum and minimum values of quadratic functions* 1) The first step is to identify what value the question is asking for and to write it in a function form. They're asking for the maximum area of a rectangular playground ($length \times width$) whose total fencing must equal $400$ feet after dividing the playground into $2$. Let us assume that one pair of the playground's sides measures $x$ and the other measures $y$. Since we don't know which variable represents length and which represents width, we can arbitrarily assign the new fence between either pair. That way, we can say that: $$TotalFence=400=3x+2y$$ Since we need to write a function in terms of just one variable, we can solve for $y$ in the following manner: $$400=3x+2y$$ $$400−3x=2y$$ $$\frac{400 - 3x}{2} = \frac{2y}{2}$$ $$200−\frac{3}{2}x=y$$ which means that one side of the playground should measure $x$ and the perpendicular side should measure ($200−\frac{3}{2}x$). The area of the playground (which is the real value that the question is seeking), can now be represented as: $$Area=(x)(200−\frac{3}{2}x)=200x−\frac{3}{2}x^{2}$$ 2) We can re-write the equation we made for area into a quadratic function with form $f(x)=ax2+bx+c$: $$f_{Area}(x)=−\frac{3}{2}x^{2}+200x+0$$ 3) Since we know that, for quadratic equations with $a<0$, we can find the maximum value at $x=\frac{-b}{2a}$, we can determine the following: $$x_{maximum}=\frac{-b}{2a}=\frac{-200}{-2(3/2)}=\frac{200}{3} \approx 66.67$$ 4) With the value of $x$ at hand, we can now determine the maximum area for the function we've established: $$f_{maximum}(x)=f(\frac{200}{3})=200(\frac{200}{3})−\frac{3}{2}(\frac{200}{3})^{2}$$ $$f(100)=\frac{40,000}{3}−\frac{3}{2}(\frac{40000}{9})$$ $$f(100) = \frac{40,000}{3} - \frac{20,000}{3} = \frac{20,000}{3} \approx 6,666.67$$ 5) Now that we have the value of the maximum area, the only value missing is the remaining pair of the playground's sides which we previously denominated as $y$, and which we can calculate using the formula we derived in step 1: $$y=200−\frac{3}{2}x_{maximum}$$ $$y=200−\frac{3}{2}(\frac{200}{3})=200−100=100$$
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