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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



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


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



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