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

Solution: Here, $x+y=35$.
$\therefore x=35-y$
Let $P=x^2{y}^5$
Then, $P={\left(35-y\right)}^2{\left(y\right)}^5$
Now, for $P$ to be maximum, $\frac{dP}{dy}$ should be $0$.
$\therefore y^5\left(2\left(35-y\right)\right)\left(-1\right)+{\left(35-y\right)}^2\left(5y^4\right)=0$
$\Rightarrow y^4\left(35-y\right)\left[-2y+5\left(35-y\right)\right]=0$
$\Rightarrow y^4\left(35-y\right)\left(175-7y\right)=0$
$\Rightarrow y=0\text{and}y=35\text{and}y=25$
When $y=0,x=35$, then our product is $0$ that is minimum.
When $y=35,x=0$, then our product is $0$ that is minimum.
When $y=25,x=10$, then our product is will be maximum.
So, $y=25\text{and}x=10$ are the required values.