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