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: 65

Answer

Length of plot $=150ft$. Width of plot $=300ft$. Maximum area of plot $=45,000f{{t}^{2}}$

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 $600ft$. $2x+y=600$. Solve for $y$. $y=600-2x$. Area of the plot, $A$ , can be written as $A=xy$. Put $y=600-2x$. $\begin{align} & A=x\left( 600-2x \right) \\ & =-2{{x}^{2}}+600x \end{align}$. And compare with the quadratic function, $f\left( x \right)=a{{x}^{2}}+bx+c$. It gives $a=-2,\ b=600$. 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: $\begin{align} & -\frac{b}{2a}=-\left( \frac{600}{-4} \right) \\ & =150. \end{align}$ Put $x=150$ in equation for $y$. $\begin{align} & y=600-2x \\ & =600-2\left( 150 \right) \\ & =300. \end{align}$ Hence, for the area of the plot to be maximum the length of the plot would be $300ft$ and width would be $150ft$. The area of the plot would be $\begin{align} & A=xy \\ & =\left( 150\times 300 \right)f{{t}^{2}} \\ & =45,000f{{t}^{2}}. \end{align}$
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