Question

The section of a window cosists of a rectangle surmounted by and equilateral triangle. If the perimeters be given as 16m, find the dimensions of the window in order that the maximum amount of light may be admitted.

• A. $2,5$
• B. $3,3.5$
• C. $3.75,2.375$
• D. $4,2$

Answer 1

Answer: (C)

$3.75,2.375$

Perimeter of window $P=2y+3x=16$

$\Rightarrow y=\frac{16-3x}{2}$ ....(1)

Area $A=xy+\frac{\sqrt{3}}{4}{x}^2$

$=\frac{\sqrt{3}}{4}{x}^2+x\left(\frac{16-3x}{2}\right)$

$A=8x+(\frac{\sqrt{3}}{4}-\frac{3}{2})x^2$

$\frac{dA}{dx}=8+(\frac{\sqrt{3}}{4}-\frac{3}{2})2x$

For maxima or minima,

$\frac{dA}{dx}=0$

$\Rightarrow 4-\frac{(6-\sqrt{3})}{4}x=0.$

$\therefore x=\frac{16}{6-\sqrt{\left(3\right)}}=\frac{16(6+\sqrt{3})}{36-3}=\frac{16(6+1.73)}{33}$

$=\frac{16\left(7.73\right)}{33}=\frac{123.68}{33}$

$\Rightarrow x=3.75$ nearly.

Now, $\frac{d^{2}A}{d{x}^2}=2(\frac{\sqrt{3}}{4}-\frac{3}{2})<0$

Hence A is maximum.

By (1),

$y=2.375$