Question

A box of constant volumec is to be twice as long is it is wide. The material on the top and four sides cost three times as much per square metre as thatin the bottom. What are the most economical dimensions?

• A. ${\left[\frac{3c}{16}\right]}^{1/3},2{\left[\frac{3c}{16}\right]}^{1/3}$ and ${\left[\frac{32c}{81}\right]}^{1/3}.$
• B. ${\left[\frac{9c}{16}\right]}^{1/3},2{\left[\frac{9c}{16}\right]}^{1/3}$ and ${\left[\frac{32c}{81}\right]}^{1/3}.$
• C. ${\left[\frac{9c}{16}\right]}^{1/3},2{\left[\frac{9c}{16}\right]}^{1/3}$ and ${\left[\frac{32c}{9}\right]}^{1/3}.$
• D. ${\left[\frac{81c}{16}\right]}^{1/3},2{\left[\frac{81c}{16}\right]}^{1/3}$ and ${\left[\frac{32c}{81}\right]}^{1/3}.$

Answer 1

Answer: (B)

Let the breadth be $x$, length be $2x$ and height be $h$

$V=x.2x.h$

$\Rightarrow c=2x^{2}h$ ....(1)

Area of bottom $=2x^2$ =Area of top

Area of sides $=2xh+2xh+xh+xh=6xh.$

If R rupees be the cost of material for bottom then for the top and sides is 3R.

$\therefore E=R\left(2x^2\right)+3R(2x^2+6xh)$

$\Rightarrow E=R(8x^2+18xh)$

or $E=R(8x^2+18x\frac{c}{2x^{{}^2}})$

$\Rightarrow E=R(8x^2+\frac{9c}{x})$

where R and c are constants.

$\frac{dE}{dx}=R(16x-\frac{9c}{x^2})$

For maximum or minimum,

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

$\therefore x={\left(\frac{9c}{16}\right)}^{1/3}$

Also, $\frac{d^{2}E}{d{x}^2}=R(16+\frac{18c}{x^3})=+ive$

and hence minimum.

$\therefore$ dimensions are

${\left[\frac{9c}{16}\right]}^{1/3},2{\left[\frac{9c}{16}\right]}^{1/3}$ and ${\left[\frac{32c}{81}\right]}^{1/3}.$