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

Let the given two numbers be $x$ and $y$. Then,

\[x\text{ }+\text{ }y\text{ }=\text{ }15\text{ }\ldots ..\text{ }\left( 1 \right)\]

\[y\text{ }=\text{ }\left( 15\text{ }-\text{ }x \right)\]

Now we have, \[z\text{ }=\text{ }{{x}^{2}}~{{y}^{3}}\]

\[z\text{ }=\text{ }{{x}^{2}}~{{\left( 15\text{ }-\text{ }x \right)}^{3}}~\left( from\text{ }equation\text{ }1 \right)\]

RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 18