Question

A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening.$$$$

Answer 1

Hint: The maximum light may only come through maximum area. We take the length of the rectangle as $x$ and breadth as $y$. We use the value of the given semi-perimeter to express $y=f\left(x\right)$. We put $y$ in the expression for the total area of the window and maximize it to find $x$ as a critical point using the second derivative test. We then find $y.$$$$$

Complete step-by-step solution
We know from the second derivative test that the function $f\left(x\right)$ will have a local maxima at critical point $x=c$ if $f^{{}^{″}}\left(c\right)<0$. The critical points are the solutions of $f^{{}^{\prime }}\left(x\right)=0$.$$$$

We are given that a window is in the form of a rectangle surmounted by a semicircular opening. If we want maximum light to pass through the opening we want the light passed through the entire area made by a rectangular window and the semi-circular opening. $$$$
Let us assume that the length of the rectangle is $x$ and breadth is $y$. So the radius of the semi-circular opening is ${\displaystyle \frac{x}{2}}.$ It is given in the question that the perimeter of the entire window is 10m. We have
$$\begin{array}{rl}& 2(x+y)+\pi \left({\displaystyle \frac{x}{2}}\right)-x=10\\ & \Rightarrow x+2y+{\displaystyle \frac{\pi x}{2}}=10\\ & \Rightarrow y=5-x({\displaystyle \frac{1}{2}}+{\displaystyle \frac{\pi }{4}})...(1)\end{array}$$
The total area of the window is the sum of the areas of rectangle and semi-circle. We have,
$$A=xy+{\displaystyle \frac{1}{2}}\pi {\left({\displaystyle \frac{x}{2}}\right)}^2$$
We put the obtained values of $y$ in the above equation to get,
$$\begin{array}{rl}& A=x(5-x({\displaystyle \frac{1}{2}}+{\displaystyle \frac{\pi }{4}}))+{\displaystyle \frac{1}{2}}\pi {\left({\displaystyle \frac{x}{2}}\right)}^2\\ & \Rightarrow A=5x-{\displaystyle \frac{1}{2}}{x}^2-{\displaystyle \frac{\pi }{4}}{x}^2+{\displaystyle \frac{\pi }{8}}{x}^2\\ & \Rightarrow A=5x-{\displaystyle \frac{1}{2}}{x}^2-{\displaystyle \frac{\pi }{8}}{x}^2\end{array}$$
We have to maximize the area. Let us differentiate $A$ once with respect to $x$ and the equate to 0 so that we can find critical points where $A$ can have a maxima. We have,
$$\begin{array}{rl}& {\displaystyle \frac{dA}{dx}}=5-x-{\displaystyle \frac{\pi x}{4}}=0\\ & \Rightarrow x={\displaystyle \frac{5}{1+{\displaystyle \frac{\pi }{4}}}}={\displaystyle \frac{20}{\pi +4}}\end{array}$$
We have obtained only one critical point $x={\displaystyle \frac{20}{\pi +4}}$. We find the value of the second derivative of $A$ at $x={\displaystyle \frac{20}{\pi +4}}$.
$$\begin{array}{rl}& {\displaystyle \frac{d^{2}A}{dx}}={\displaystyle \frac{d}{dx}}\left({\displaystyle \frac{dA}{dx}}\right)\\ & \Rightarrow {\displaystyle \frac{d^{2}A}{dx}}={\displaystyle \frac{d}{dx}}(5-x-{\displaystyle \frac{\pi x}{4}})=-1-{\displaystyle \frac{\pi }{4}}\end{array}$$
We see that ${\displaystyle \frac{d^{2}A}{dx}}<0$ for all $x\in R$ including $x={\displaystyle \frac{20}{\pi +4}}$. So $A$ will have a maxima at $x={\displaystyle \frac{20}{\pi +4}}$. We put $x$ in equation (1) and get,
$$\begin{array}{rl}& y=5-\left({\displaystyle \frac{20}{\pi +4}}\right)({\displaystyle \frac{1}{2}}+{\displaystyle \frac{\pi }{4}})=5-\left({\displaystyle \frac{20}{\pi +4}}\right)\left({\displaystyle \frac{2+\pi }{4}}\right)\\ & =5-5{\displaystyle \frac{\pi +2}{\pi +4}}=5\left({\displaystyle \frac{\pi +2-\pi -4}{\pi +4}}\right)=5{\displaystyle \frac{2}{\pi +4}}={\displaystyle \frac{10}{\pi +4}}\end{array}$$
So the dimension of the window to admit maximum light through the whole opening is length ${\displaystyle \frac{20}{\pi +4}}$m and breadth ${\displaystyle \frac{10}{\pi +4}}$m.$$$$

Note: We need to be careful that the perimeter of the window does not include the diameter of the semicircle. We note that if the second derivative would have $f^{{}^{″}}\left(c\right)=0$ we would have done a third derivative test and so on. The function $f\left(x\right)$ has a maxima at $x=c$ if $f^{n+1}\left(c\right)<0$ and $f^{{}^{\prime }}\left(c\right)=f^{{}^{″}}\left(c\right)=...=f^n\left(c\right)=0$