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  • Explain solution RD Sharma class12 chapter Maxima and Minima exercise 17.5 question 24 maths.

Explain solution RD Sharma class12 chapter Maxima and Minima exercise 17.5 question 24 maths.

Answers (1)

l =12

Hint: For maxima or minima f'(x) must be zero f'(x) = 0

Given: Let l,b,v be length, breadth and volume

Solution:

2(l+b)=36

l =18-b                 .....(1)

V= \pi l^2 b

V= \pi (18-b)^2 b                    ..... from (1)

\begin{aligned} &=\pi\left(324 b+b^{3}-36 b^{2}\right) \\ &\frac{d V}{d b}=\pi\left(324+3 b^{2}-72 b\right) \\ &\frac{d V}{d b}=0 \end{aligned}

\begin{aligned} &\pi\left(324+3 b^{2}-72 b\right)=0 \\ &324+3 b^{2}-72 b=0 \\ &b^{2}-24 b+108=0 \\ &b^{2}-6 b-18 b+108=0 \\ &(b-6)(b-18)=0 \\ &b=6,18 \end{aligned}

Now,

\begin{aligned} &\frac{d^{2} V}{d b^{2}}=\pi(6 b-72) \\ &\text { At } b=6 \\ &\frac{d^{2} V}{d b^{2}}=\pi(36-72)=-36 \pi<0 \\ &\text { At } b=18 \\ &\frac{d^{2} V}{d b^{2}}=\pi(18 \times 6-72)=36 \pi>0 \end{aligned}

Substitute the value of  b in (1)

l=18-6=12

So, the volume is maximum when a = 12cm and b = 6cm 

 

 

 

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