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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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