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

Question:

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

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

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