Question

Find two positive numbers x and y such that x+y = 60 and $x{y}^3$ is maximum.

Answer 1

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
$\frac{dP}{dy}=180y^2-4y^3$
and $\frac{d^{2}P}{d{y}^2}=360y-12y^2$
For maxima, we must have $\frac{dP}{dy}=0$
$\Rightarrow 180y^2-4y^3=0\Rightarrow 4y^2(45-y)=0\Rightarrow y=45$
But $y\ne 0$, so y=45
At $y=45{\left(\frac{dP}{d{y}^2}\right)}_{y=45}=360\times 45-12\times (45{)}^2=16200-24300=-8100<0$
P has a local maxima at y=45
Therefore, 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.