Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 2 - Section 2.2 - Quadratic Functions - Exercise Set - Page 331: 66


Length of plot $=50ft$ Width of plot $=100ft$ Maximum area of plot $=5,000f{{t}^{2}}$

Work Step by Step

Let us assume that the length of the plot be $y$ and breadth of the plot along the river be $x$. The plot is to be fenced from three sides and is to be left along the riverside. The total available length of fence is $200ft$. $2x+y=200$. Solve for $y$. $y=200-2x$. Area of the plot, $A$ , can be written as $A=xy$. Put $y=200-2x$. $\begin{align} & A=x\left( 200-2x \right) \\ & =-2{{x}^{2}}+200x \end{align}$. And compare with the quadratic function, $f\left( x \right)=a{{x}^{2}}+bx+c$. It gives $a=-2,b=200$. Since $a<0$ , the given quadratic function has a maximum. The value at which the function gives a maximum value is calculated by evaluating $-\frac{b}{2a}$ which is shown as follows: $\begin{align} & -\frac{b}{2a}=-\left( \frac{200}{-4} \right) \\ & =50 \end{align}$. Put $x=50$ in equation for $y$. $\begin{align} & y=200-2x \\ & =200-2\left( 50 \right) \\ & =100. \end{align}$ Hence, for the area of the plot to be maximum the length of the plot would be $100ft$ and width would be $50ft$. The area of the plot would be $\begin{align} & A=xy \\ & =\left( 50\times 100 \right)f{{t}^{2}} \\ & =5,000f{{t}^{2}}. \end{align}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.