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

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

Answers (1)

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} 

Posted by

infoexpert24

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