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  • Provide solution for RD Sharma maths class12 Chapter Maxima and Minima exercise 17.5 question 12.

Provide solution for RD Sharma maths class12 Chapter Maxima and Minima exercise 17.5 question 12.

Answers (1)

 432 cm2

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

Given: From the question

V(x)=x(18-2x)^2

Solution:

Let the square side be  cutoff be xcm. Then, the length and breadth of the box will be (18-2x)cm each and height of the box will be xcm

Volume of the box V(x)=x(18-2x)^2

\begin{aligned} &V^{\prime}(x)=(18-2 x)^{2}-4 x(18-2 x) \\ &=(18-2 x)^{2}(18-2 x-4 x) \\ &=(18-2 x)^{2}(18-6 x) \end{aligned}

\begin{aligned} &=12(9-x)(3-x) \\ &V^{\prime \prime}(x)=12(-(9-x)-(3-x)) \\ &=-12(9-x+3-x)=-24(6-x) \end{aligned}

For maxima or minima V'(x) =0

12(9-x)(3-x)=0

x=9 or x =3

if x=9, then l and b will become 0

So, x\neq 9

if x =3

Now, V''(3)= -24(6-3)=-72<0

 x= 3 is the point of maximum

V(3)= 3(18-6)^3=3 \times 144

=432 \; cm^2

 

 

Posted by

infoexpert24

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