Chapter 7 - Section 7.4 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 851: 58

The length and width are 12 feet by 8 feet

Work Step by Step

Let us assume $l$ be the length and $w$ be the width of the rectangle such that: \begin{align} & 2l+2w=40\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{I} \right) \\ & lw=96\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{II} \right) \end{align} Divide the equation (I) by $2$ and the value of $l$ is, \begin{align} & l+w=20 \\ & l=20-w\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{III} \right) \\ \end{align} Substitute the value of (I) in equation (II). Then, \begin{align} & \left( 20-w \right)w=96 \\ & 20w-{{w}^{2}}=96 \\ & {{w}^{2}}-20w+96=0 \end{align} Now, solve for $w$. Therefore, \begin{align} & \left( {{w}^{2}}-12 \right)w-8w+96=0 \\ & \left( w-12 \right)\left( w-8 \right)=0 \\ & w=12,8 \end{align} Substitute the value of $w=11,7$ into equation (III): If $w=12$ then, \begin{align} & l=20-12 \\ & =8 \end{align} If $w=8$ then, \begin{align} & l=20-8 \\ & =12 \end{align} Thus, the dimensions of the rectangle are 12 feet by 8 feet.

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