Question

A normal window has the shape of a rectangle surmounted by a semicircle. (thus the diameter of the semicircle is equal to the width of the rectangle.) if the perimeter of the window is $16\mathrm{ft}$, find the value of $x$ so that the greatest possible amount of light is admitted. (give your answer correct to two decimal places.)

Answer 1

Step-1: Framing the relation between $\mathit{x}$ and $\mathit{y}$:

The perimeter of the window is $16\mathrm{ft}$.

The perimeter of the window is the sum of perimeter of the rectangle and perimeter of the semicircle minus width of the rectangle. (normal window has the shape of a rectangle surmounted by a semicircle)

The perimeter of the rectangle $P$ is calculated by $P=2\left(x+y\right)$ where $x$ represents the width of the rectangle and $y$ represents the length of the rectangle.

The width of the rectangle is equal to the diameter of the semicircle.

The perimeter of the semicircle $P\text{'}$ is calculated by $P\text{'}=\pi \left(\frac{x}{2}\right)+2\left(\frac{\mathrm{x}}{2}\right)$ where $\frac{x}{2}$ represents the radius of the semicircle.

$\begin{array}{rcl}16& =& P+P\text{'}-x\\ 16& =& 2\left(x+y\right)+\pi \left(\frac{x}{2}\right)+2\left(\frac{x}{2}\right)-x\\ 16& =& 2x+2y+\frac{\mathrm{\pi x}}{2}+x-x\\ 16& =& 2x+2y+\frac{\mathrm{\pi x}}{2}\\ 2y& =& 16-2x-\frac{\mathrm{\pi x}}{2}\\ y& =& 8-x-\frac{\mathrm{\pi x}}{4}\end{array}$

Step-2: Determine the area of the window $\mathit{A}$:

The area of the window is the sum of area of the rectangle and area of the semicircle.

The area of the rectangle $A\text{'}$ is calculated by $A\text{'}=xy$.

The area of the semicircle $A\text{'}\text{'}$ is calculated by $A\text{'}\text{'}=\frac{\pi }{2}·{\left(\frac{x}{2}\right)}^2$.

$\begin{array}{rcl}A& =& A\text{'}+A\text{'}\text{'}\\ & =& xy+\frac{\mathrm{\pi }}{2}·{\left(\frac{x}{2}\right)}^2\\ & =& x\left(8-x-\frac{\mathrm{\pi x}}{4}\right)+\frac{{\mathrm{\pi x}}^2}{8}\\ & =& 8x-x^2-\frac{{\mathrm{\pi x}}^2}{4}+\frac{{\mathrm{\pi x}}^2}{8}\\ & =& 8x-x^2-\frac{{\mathrm{\pi x}}^2}{8}\end{array}$

Step-3: Determine the value of $\mathit{x}$:

Differentiate $A$ with respect to $x$.

$\frac{dA}{dx}=8-2x-\frac{\mathrm{\pi x}}{4}$

For critical point, $\frac{dA}{dx}=0$.

$\begin{array}{rcl}8-2x-\frac{\mathrm{\pi x}}{4}& =& 0\\ 8& =& 2x+\frac{\mathrm{\pi x}}{4}\\ x& =& \frac{8}{2+{\displaystyle \frac{\mathrm{\pi }}{4}}}\\ & =& \frac{32}{8+\mathrm{\pi }}\\ & \approx & 2.86\mathrm{ft}\end{array}$

Area expression is in downward parabola so maximum value will occur at $x=2.86$.

Hence, the value of $\mathit{x}$ is $\mathbf{2}\mathbf{.}\mathbf{86}\mathbf{}\mathbf{ft}$.