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 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
$P=(60-y)y^3\Rightarrow P=60y^3-y^4$
On differentiating twice w.r.t. $y$, we get
and $\frac{dP}{dy}=180y^2-4y^3$
$\frac{d^{2}P}{d{y}^2}=360y-12y^2$
For maxima, we must have $\frac{dP}{dy}=0$.
$\therefore 180y^2-4y^3=0$
$\Rightarrow 4y^2(45-y)=0$
$\Rightarrow y=0,45$
But $y\ne 0$, so $y=45$
At $y=45,{\left(\frac{d^{2}P}{d{y}^2}\right)}_{y-45}=360\times 45-12\times (45{)}^2$
$=16200-24300=-8100<0$
$\Rightarrow P$ 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 .