#### Question

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

#### Solution

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

\[x + y = 15 . . . (1)\]

\[\text { Now,} \]

\[ z = x^2 y^3 \]

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

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

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

\[\frac{dz}{dx} = 0\]

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

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

\[ \Rightarrow 30x - 2 x^2 = 3 x^2 \]

\[ \Rightarrow 30x = 5 x^2 \]

\[ \Rightarrow x = 6 \text { and }y = 9\]

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

\[\text { At x } = 6: \]

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

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

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

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