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 332: 68


Length of plot $\ =20yd$ Width of plot $\ =20yd$ Maximum area of plot $\ =400y{{d}^{2}}$

Work Step by Step

Let us assume the length of the plot be $x$ and breadth of the plot along the river be $y$. The plot is to be fenced from three sides and is to be left along the riverside. The total available length of fence is $80$ yards. $2x+2y=80$. Solve for $y$. $\begin{align} & 2y=80-2x \\ & y=40-x. \end{align}$ Area of the plot, $A$ , can be written as $A=xy$. Put $y=40-x$. $\begin{align} & A=x\left( 40-x \right) \\ & =-{{x}^{2}}+40x. \end{align}$ And compare with the quadratic function, $f\left( x \right)=a{{x}^{2}}+bx+c$. It gives $a=-1,\ b=40.$ 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: $\begin{align} & -\frac{b}{2a}=-\left( \frac{40}{-2} \right) \\ & =20. \end{align}$ Put $x=20$ in equation for $y$. $\begin{align} & y=40-x \\ & =40-20 \\ & =20. \end{align}$ Hence, for the area of the plot to be maximum the length of the plot would be $20$ yards and width would be $20$ yards. The area of the plot would be $\begin{align} & A=xy \\ & =\left( 20\times 20 \right) \\ & =400y{{d}^{2}} \end{align}$
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