Question

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

• A. $13$, $47$
• B. $28$, $32$
• C. $12$, $48$
• D. $15$, $45$

Answer 1

Answer: (D)

Let $P=x{y}^3$
Given, $x+y=60$
$\Rightarrow x=60-y$
On putting this value in $P=x{y}^3$, we get
$P=\left(60-y\right)y^3$
$\Rightarrow P=60y^3-y^4$
On differentiating w.r.t. $y$, we ge
$\frac{dP}{dy}=180y^2-4y^3$
$\Rightarrow \frac{d^{2}P}{d{y}^2}=360y-12y^2$
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 $y=45$, ${\left(\frac{d^{2}P}{d{y}^2}\right)}_{y=45}=360\times 45-12\times {\left(45\right)}^2$
$=16200-24300=-8100<0$
$\Rightarrow P$ has a local maxima at $y=45$.
By second derivative test, $y=45$ is a point of local maxima of $P$. Thus, function $x{y}^3$ is maximum when $y=45$ and $x=60-45=15$.