Chapter 3, Problem 44RE

To determine

Largest area $A$ of that cam enclosed by the rectangle.

Answer to Problem 44RE

Largest area $A$ of that cam enclosed by the rectangle is $25m^2$ .

Explanation of Solution

Given information:

  $y=10-x,x>0$

Calculation:

Initially lets draw a $OABC$ , having a vertex $O$ in the origin, a vertex $A$ on the $x-$ axis, a vertex $B$ on the line $y=10-x$ and a vertex $C$ on the $y-$ axis.

Coordinates of the point $A$ as $A\left(x,0\right)$

Because $B$ belongs to the line $y=10-x$ , the coordinates of the point $B$ are:

  $B\left(x,10-x\right)$

the dimensions of $OABC$ rectangle are:

Width $=x$

length $=10-x$

Now one can compute the rectangle’s area as a function of $x$ :

  $\begin{array}{l}A=length.width\\ A\left(x\right)=\left(10-x\right)x\\ A\left(x\right)=x^2+10x\end{array}$

Because in the formula of $A\left(x\right)$ , the coefficient of $x^2$ is negative, the parabola has a maximum for the value of $x$ .

  $\begin{array}{l}{x}_{\max }=-\frac{10}{2\left(-1\right)}\\ x_{\max }=\frac{10}{2}\\ x_{\max }=5\end{array}$

Now. One can compute the maximum area by substituting the value of $x_{\max }$ in the parabola’s equation:

  $\begin{array}{l}{A}_{\max }=A\left(5\right)=-5^2+10.5\\ A_{\max }=-25+50\\ A_{\max }=25\end{array}$

Thus, area is $25m^2$ .