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# Divide 64 into Two Parts Such that the Sum of the Cubes of Two Parts is Minimum. - CBSE (Science) Class 12 - Mathematics

#### Question

Divide 64 into two parts such that the sum of the cubes of two parts is minimum.

#### Solution

$\text { Suppose 64 is divided into two partsxand 64-x. Then, }$

$z = x^3 + \left( 64 - x \right)^3$

$\Rightarrow \frac{dz}{dx} = 3 x^2 + 3 \left( 64 - x \right)^2$

$\text { For maximum or minimum values of z, we must have }$

$\frac{dz}{dx} = 0$

$\Rightarrow 3 x^2 + 3 \left( 64 - x \right)^2 = 0$

$\Rightarrow 3 x^2 = 3 \left( 64 - x \right)^2$

$\Rightarrow x^2 = x^2 + 4096 - 128x$

$\Rightarrow x = \frac{4096}{128}$

$\Rightarrow x = 32$

$\text { Now, }$

$\frac{d^2 z}{d x^2} = 6x + 6\left( 64 - x \right)$

$\Rightarrow \frac{d^2 z}{d x^2} = 384 > 0$

$\text { Thus, z is minimum when 64 is divided into two equal parts, 32 and 32.}$

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Solution Divide 64 into Two Parts Such that the Sum of the Cubes of Two Parts is Minimum. Concept: Graph of Maxima and Minima.
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