A box of constant volume c is to be twice as long as it is wide.

Solution:

Let $l, b$ and $h$ be the length, breadth and height of the box, respectively.

Volume of the box $=c$

Given : $l=2 b$         $\cdots(1)$

$\Rightarrow c=l b h$

$\Rightarrow c=2 b^{2} h$

$\Rightarrow h=\frac{c}{2 b^{2}}$              .......(2)

Let cost of the material required for bottom be $K / \mathrm{m}^{2}$.

Cost of the material required for 4 walls and top $=\mathrm{Rs} 3 K / \mathrm{m}^{2}$

Total cost, $T=K(l b)+3 k(2 l h+2 b h+l b)$

$\Rightarrow T=2 K b^{2}+3 K\left(\frac{4 b c}{2 b^{2}}+\frac{2 b c}{2 b^{2}}+2 b^{2}\right)$       [From eqs. (1) and (2)]

$\Rightarrow \frac{d T}{d b}=4 K b+3 K\left(\frac{-3 c}{b^{2}}+4 b\right)$

For maximum or minimum values of $T$, we must have

$\frac{d T}{d b}=0$

$\Rightarrow 4 k b+3 K\left(\frac{-3 c}{b^{2}}+4 b\right)=0$

$\Rightarrow 4 b=3\left(\frac{3 c}{b^{2}}-4 b\right)$

$\Rightarrow 4 b=\left(\frac{9 c}{b^{2}}-12 b\right)$

$\Rightarrow 4 b=\frac{9 c-12 b^{3}}{b^{2}}$

$\Rightarrow 4 b^{3}=9 c-12 b^{3}$

$\Rightarrow 16 b^{3}=9 c$

$\Rightarrow b=\left(\frac{9 c}{16}\right)^{\frac{1}{3}}$

Now,

$\frac{d^{2} T}{d b^{2}}=4 K+3 K\left(\frac{6 c}{b^{3}}+4\right)$

$\Rightarrow \frac{d^{2} T}{d b^{2}}=4 K+3 K\left(\frac{6 c}{9 c} \times 16+4\right)$

$\Rightarrow K\left(4+3 \times \frac{44}{3}\right)$

$\Rightarrow 48 K>0$

$\therefore$ Cost is minimum when $b=\left(\frac{9 c}{16}\right)^{\frac{1}{3}}$.

Substituting $b=\left(\frac{9 c}{16}\right)^{\frac{1}{3}}$ in eq. (1) and eq. (2)

$\Rightarrow l=2\left(\frac{9 c}{16}\right)^{\frac{1}{3}}$

$h=\frac{c}{2 b^{2}}$

$\Rightarrow h=\frac{c}{2\left(\frac{9 c}{16}\right)^{\frac{2}{3}}}$

$\Rightarrow h=\left(\frac{32 c}{81}\right)^{\frac{1}{3}}$

Thus, the most economic dimensions of the box are $l=2\left(\frac{9 c}{16}\right)^{\frac{1}{3}}, b=\left(\frac{9 c}{16}\right)^{\frac{1}{3}}$ and $h=\left(\frac{32 c}{81}\right)^{\frac{1}{3}}$.