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