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  • Provide solution for RD Sharma maths class12 Chapter Maxima and Minima exercise 17.5 question 13.

Provide solution for RD Sharma maths class12 Chapter Maxima and Minima exercise 17.5 question 13.

Answers (1)

5cm

Hint: For maxima and minima we must have f'(x) =0

Given: For the question,

V= (45-2x)(24-2x)x

Solution: Let the side of the square to be xcm is cutoff.

Then, the l and b, h of the loon will be (45-x), (25-x),xcm

Volume of the box V = (45-x) (25-x)x

\frac{d V}{d x}=(45-2 x)(24-2 x)-2 x(45-2 x)-2 x(24-2 x)

\frac{d V}{d x}=0

\begin{aligned} &(45-2 x)(24-2 x)-2 x(45-2 x)-2 x(24-2 x)=0 \\ &4 x^{2}+1080-138 x-48 x+4 x^{2}+4 x^{2}-90 x=0 \\ &12 x^{2}-276 x+1080=0 \end{aligned}

Divide by 12

\begin{aligned} &x^{2}-23 x+90=0 \\ &x^{2}-18 x-5 x+90=0 \\ &x(x-15)-5(x-18)=0 \\ &x-18=0 \text { or } x-5=0 \\ &x=18 \text { or } x=5 \end{aligned}

\begin{aligned} &\frac{d^{2} V}{d x^{2}}=24 x-276 \\ &\frac{d^{2} V}{d x^{2}}=120-276=-156<0 \\ &\frac{d^{2} V}{d x^{2}}_{x=18}=432-276=156>0 \end{aligned}

Thus volume of the box is max when x = 5cm.

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