A wire of length 20 units is divided into two parts such that the product of one part and cube of the other part is maximum, then product of these parts is

Let \(x\) be the one part and \(y\) be the other part.
We have \(\mathrm{x}+\mathrm{y}=20 \Rightarrow \mathrm{y}=20-\mathrm{x}\)
As per condition given, we write
\(
\begin{aligned}
& \mathrm{f}(\mathrm{x})=(20-\mathrm{x}) \mathrm{x}^3 \\
& =20 \mathrm{x}^3-\mathrm{x}^4 \\
& \therefore \mathrm{f}^{\prime}(\mathrm{x})=60 \mathrm{x}^2-4 \mathrm{x}^3
\end{aligned}
\)
When \(\mathrm{f}^{\prime}(\mathrm{x})\), we get
\(
\begin{aligned}
& 4 x^2(15-x)=0 \Rightarrow x=0,15 \\
& f^{\prime}(x)=120 x-12 x^2 \\
& {\left[f^{\prime}(x)\right]_{x=15}=(120)(15)-(12)(15)^2=-900 < 0}
\end{aligned}
\)
\(\therefore \mathrm{f}[\mathrm{x}]\) is maximum when \(\mathrm{x}=15\).
\(
\therefore \mathrm{y}=5 \Rightarrow \mathrm{xy}=(15)(5)=75
\)