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Two positive numbers $\mathrm{x}$ and $\mathrm{y}$ such that $(x+y)=60$ and $x y^3$ is maximum, then the numbers are respectively
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Verified Answer
The correct answer is:
45,15
$x+y=60 \Rightarrow x=60-y$
Now let $z=x y^3=(60-y) y^3=60 y^3-y^4$
$$
\Rightarrow \frac{d z}{d y}=180 y^2-4 y^3=4 y^2(45-y)
$$
i.e., $x y^3$ is maximum $y=45$ and $x=60-45=15$
Now let $z=x y^3=(60-y) y^3=60 y^3-y^4$
$$
\Rightarrow \frac{d z}{d y}=180 y^2-4 y^3=4 y^2(45-y)
$$
i.e., $x y^3$ is maximum $y=45$ and $x=60-45=15$
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