IIT-JEE 1991Question 1
A window of perimeter $P$ (including the base of the arch) is in the form of a rectangle surmounted by a semi circle. The semi-circular portion is fitted with coloured glass while the rectangular part is fitted with clear glass transmits three times as much light per square meter as the coloured glass does. What is the ratio for the sides of the rectangle so that the window transmits the maximum light?
Differentiation and its Application
Answerheight : width $=(6+\pi):6$ (width $=2r=\dfrac{6P}{24+5\pi}$, height $=\dfrac{(6+\pi)P}{24+5\pi}$)
Solution
Concept. Let the rectangle be $2r$ wide and $h$ high. Write the light in terms of $r$ (using the perimeter) and maximise. Formulas used. (1) Perimeter with the base of the arch: $P=2h+2r+2r+\pi r=2h+(4+\pi)r$(2) Light (coloured glass $=1$ unit per m$^2$): $L=3\cdot2rh+\frac12\pi r^2$Step 1 - The window.Step 2 - $L$ in $r$. By (1), $2rh=r\big(P-(4+\pi)r\big)$, so by (2) $$L(r)=3Pr-3(4+\pi)r^2+\frac{\pi r^2}{2}.$$ Step 3 - Maximise. $$L'(r)=3P-(24+5\pi)r=0\ \Longrightarrow\ r=\frac{3P}{24+5\pi}.$$ $L''<0$, so this is the maximum. Step 4 - Sides. Width $2r=\frac{6P}{24+5\pi}$ and $$h=\frac{P-(4+\pi)r}{2}=\frac{(6+\pi)P}{24+5\pi}.$$ So height : width $=(6+\pi):6$. Answer: height : width $=(6+\pi):6$ (width $=2r=\dfrac{6P}{24+5\pi}$, height $=\dfrac{(6+\pi)P}{24+5\pi}$)