Chapter 3, Problem 42RE

To determine

Dimensions of the rectangle that maximizes its area.

Answer to Problem 42RE

Dimension of rectangle that maximizes its area are $l=50,x=50$

Explanation of Solution

Given information:

Length $=100feet$

Calculation:

Let, the length and width of the rectangle be $l$ and $x$ .

So, radius of semicircle is $\frac{x}{2}$ .

So we have,

  $\begin{array}{l}\frac{x}{2}+l+\frac{x}{2}=100\\ l+x=100\\ l=100-x\end{array}$

The area of $A$ of rectangle is $l\times x$ .

Substituting the value of $l$ we have,

  $\begin{array}{l}A=l\times x\\ =\left(100-x\right)x\\ =100-x^2\end{array}$

A special window in the shape of a rectangle with semicircles at each end is to be constructed so outside dimension are $100ft$ in the length.

For getting maximum value of area, take the derivative of $A$ with respect to $x$ and set it to zero.

  $\begin{array}{l}\frac{dA}{dx}=0\\ 100-2x=0\\ -2x=100\\ x=50\end{array}$

That is why,

  $\begin{array}{l}l=100-50\\ =50\end{array}$

Hence, dimension of rectangle that maximumizes its area are

  $\begin{array}{l}l=50\\ x=50\end{array}$