#### Provide solution for RD Sharma maths class12 Chapter Maxima and Minima exercise 17.5 question 39.

$h=\left(\frac{32 c}{81}\right)^{\frac{1}{3}}, b=\left(\frac{9 c}{16}\right)^{\frac{1}{3}}, a=\left(\frac{9 c}{16}\right)^{\frac{1}{2}}$

Hint: For maximum or minimum value of T, we must have $\frac{dT}{db}=0$

Given:

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

Volume of the box = c

Solution:

l = 2b             ....(1)

c = lbh

$c= 2b^2h$

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

Let the cost of material required to the bottom be $k/m^2$.

Cost of material required for 4 wall stand top = $Rs 3k/m^2$

Total cost , $T=k(lb)+3k(2lh +2bh + lb)$

$T=2kb^2+3k\left ( \frac{4bc}{2b^2}+\frac{2bc}{2b^2}+b^2 \right )$                [From (1) and (2)}

\begin{aligned} &\frac{d T}{d b}=4 k b+3 k\left(\frac{-3 c}{b^{2}}+4 b\right) \\ &\frac{d T}{d b}=0 \\ &4 k b+3 k\left(\frac{-3 c}{b^{2}}+4 b\right)=0 \\ &4 b=3\left(\frac{3 c}{b^{2}}+4 b\right) \end{aligned}

\begin{aligned} &4 b=\frac{9 c-12 b^{3}}{b^{2}} \\ &4 b^{3}=9 c-12 b^{3} \\ &16 b^{3}=9 c \\ &b=\left(\frac{9 c}{16}\right)^{\frac{1}{3}} \end{aligned}

\begin{aligned} &\frac{d^{2} T}{d b^{2}}=4 k+3 k\left(\frac{6 c}{b^{3}}+4\right) \\ &=4 k+3 k\left(\frac{6 c}{9 c} \times 16+4\right) \\ &=k\left(\frac{44}{3} \times 3+4\right) \\ &=48 k>0 \end{aligned}

$\therefore$ cost of the minimum when $b=\left(\frac{9 c}{16}\right)^{\frac{1}{3}}$

Sub b in eqn(1) , eqn (2)

\begin{aligned} &l=2\left(\frac{9 c}{16}\right)^{\frac{1}{3}} \\ &h=\frac{c}{2 b c^{2}}, h=\frac{c}{2\left(\frac{9 c}{16}\right)^{\frac{2}{3}}}=\left(\frac{32 c}{81}\right)^{\frac{1}{3}} \\ &\therefore l=2\left(\frac{9 c}{16}\right)^{\frac{1}{3}}, b=\left(\frac{9 c}{16}\right)^{\frac{1}{3}}, h=\left(\frac{32 c}{81}\right)^{\frac{1}{3}} \end{aligned}

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