Question

Find two positive numbers $x \space and\space y$ such that their sum is $35$ and the product $x^{2} y^{5}$ is a maximum

Answer 1

Let one number be $x$. Then, the other number is $y = (35 − x)$.

Let$p(x)=x^{2}y^{5}$.Then, we have:

$p(x)=x^{2}(35-x)^{5}$

$p'(x)=2x(35-x)^{5}-5x^{2}(35-x)^{4}$

$=x(35-x)^{4}[2(35-x)-5x]$

$=x(35-x)^{4}(70-7x)$

$=7x(35-x)^{4}(10-x)$

And$.p''(x)=7(35-x)^{4}(10-x)+7x[-35-x)^{4}-4(35-x)^{3}(10-x)]$

$=7(35-x)^{4}(10-x)-7x(35-x)^{4}-28x(35-x)^{3}(10-x)$

$=7(35-x)^{3}[(35-x)(10-x)-x(35-x)-4x(10-x)]$

$=7(35-x)^{3}[350-45x+x^{2}-35x+x^{2}-40x+4x^{2}]$

$=7(35-x)^{3}(6x^{2}-120x+350)$

Now,$p(x)=0=x=0,x=35,x=10$

When x = 35,$$ f'(x)=f(x)=0 $and$ $y = 35 − 35 = 0$. This will make the product$x^{2} y^{5}$ equal to $0$.

When $x = 0, y = 35 − 0 = 35$and the product $x^{2} y^{5}$ will be $0$.

∴ $x = 0 \space and \space x = 35$ cannot be the possible values of x. When $x = 10$, we have:
$P"(x)=7(35-10)^{3}(6\times 100-120\times10+350)$
$=7(25)^{3}(-250)<0$

∴ By second derivative test, $P(x)$ will be the maximum when $x = 10$ and $y = 35 − 10 = 25$. Hence, the required numbers are $10 \space and \space 25$.