## Precalculus (6th Edition) Blitzer

The perimeter of a rectangle is 80 feet. If the length of the rectangle is represented by x, its width can be expressed as $40-x$. The area of the rectangle, $A\left( x \right)$, expressed in the form $A\left( x \right)=a{{x}^{2}}+bx+c$, is $A\left( x \right)=-{{x}^{2}}+40x$.
The perimeter of a rectangle is 80 feet. It is assumed that the length of the rectangle is x and the width of the rectangle is y. The perimeter of the rectangle is: $\text{Perimeter}=2\left( \text{length}+\text{width} \right)$ Therefore: \begin{align} & \text{perimeter}=2x+2y \\ & 80=2\left( x+y \right) \\ & 40=x+y \\ & 40-x=y \end{align} So, the width of rectangle is $y=40-x$. The area of rectangle, \begin{align} & \text{Area}=\text{Length}\times \text{Width} \\ & =x\cdot y \end{align} The area of the rectangle is calculated as below: \begin{align} & A\left( x \right)=xy \\ & =x\left( 40-x \right) \\ & =-{{x}^{2}}+40x \end{align} So, the product of $A\left( x \right)$ is expressed in the form $A\left( x \right)=a{{x}^{2}}+bx+c$ is $-{{x}^{2}}+40x$.