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Q.
If two positive numbers $x$ and $y$ are such that $x + y =60$
and $xy^3$ is maximum, then the numbers $x$ and $y$ are respectively
Application of Derivatives
Solution:
Let $P = xy^3$
Given, $x + y = 60$
$\Rightarrow x = 60 - y$
On putting this value in $P = xy^{3}$, we get
$P = \left(60- y \right) y^{3}$
$\Rightarrow P = 60y^{3} -y^{4}$
On differentiating w.r.t. $y$, we ge
$\frac{dP}{dy} = 180y^{2} - 4y^{3}$
$\Rightarrow \frac{d^{2}P}{dy^{2}} = 360y-12y^{2}$
For maxima, we must have $\frac{dP}{dy} = 0$.
$\therefore 180y^{2} - 4y^{3} = 0$
$\Rightarrow 4y^{2}\left(45 - y\right) = 0$
$\Rightarrow y = 0$, $45$
But $y \ne 0$, so $y = 45$
At $y = 45$, $\left(\frac{d^{2}P}{dy^{2}}\right)_{y = 45} = 360 \times 45 - 12 \times \left(45\right)^{2}$
$= 16200 - 24300 = - 8100 < 0 $
$\Rightarrow P$ has a local maxima at $y = 45$.
By second derivative test, $y = 45$ is a point of local maxima of $P$. Thus, function $xy^{3}$ is maximum when $y = 45$ and $x = 60 - 45 = 15$.