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

Answers (1)


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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