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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
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Verified Answer
The correct answer is:
75
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 $\mathrm{x}=15$.
$$
\therefore \mathrm{y}=5 \Rightarrow \mathrm{xy}=(15)(5)=75
$$
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 $\mathrm{x}=15$.
$$
\therefore \mathrm{y}=5 \Rightarrow \mathrm{xy}=(15)(5)=75
$$
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