PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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# Solution for Divide 15 into Two Parts Such that the Square of One Multiplied with the Cube of the Other is Minimum. - PUC Karnataka Science Class 12 - Mathematics

#### Question

Divide 15 into two parts such that the square of one multiplied with the cube of the other is minimum.

#### Solution

$\text { Let the two numbers be x and y.Then },$

$x + y = 15 . . . (1)$

$\text { Now,}$

$z = x^2 y^3$

$\Rightarrow z = x^2 \left( 15 - x \right)^3 \left[ \text { From eq } . \left( 1 \right) \right]$

$\Rightarrow \frac{dz}{dx} = 2x \left( 15 - x \right)^3 - 3 x^2 \left( 15 - x \right)^2$

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

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

$\Rightarrow 2x \left( 15 - x \right)^3 - 3 x^2 \left( 15 - x \right)^2 = 0$

$\Rightarrow 2x\left( 15 - x \right) = 3 x^2$

$\Rightarrow 30x - 2 x^2 = 3 x^2$

$\Rightarrow 30x = 5 x^2$

$\Rightarrow x = 6 \text { and }y = 9$

$\frac{d^2 z}{d x^2} = 2 \left( 15 - x \right)^3 - 6x \left( 15 - x \right)^2 - 6x \left( 15 - x \right)^2 + 6 x^2 \left( 15 - x \right)$

$\text { At x } = 6:$

$\frac{d^2 z}{d x^2} = 2 \left( 9 \right)^3 - 36 \left( 9 \right)^2 - 36 \left( 9 \right)^2 + 6\left( 36 \right)\left( 9 \right)$

$\Rightarrow \frac{d^2 z}{d x^2} = - 2430 < 0$

$\text { Thus, z is maximum when x = 6 and y = 9 } .$

$\text { So, the required two parts into which 15 should be divided are 6 and 9 } .$

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Solution Divide 15 into Two Parts Such that the Square of One Multiplied with the Cube of the Other is Minimum. Concept: Graph of Maxima and Minima.
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