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

We have numbers $x$ and $y$ and let $P = {x^2}{y^5}$ and
$x + y = 35$ …(i) $P = {(35 - y)^2}{y^5}$ [from (i)]

Now, $\cfrac{{dP}}{{dy}} = {(35 - y)^2}(5{y^4}) + {y^5} \times 2(35 - y)( - 1)$
$= {y^4}(35 - y)\{ 5(35 - y) - 2y\} = {y^4}(35 - y)(175 - 7y)$

For maximum $P,\cfrac{{dP}}{{dy}} = 0$
$\Rightarrow {y^4}(35 - y)(175 - 7y) = 0 \Rightarrow 175 - 7y = 0$
$\Rightarrow y = 25$

Now, $\cfrac{{{d^2}P}}{{d{y^2}}} = 4(35 - y)(175 - 7y){y^3} + {y^4}( - 1)(175 - 7y) + {y^4}(35 - y)( - 7)$
$\Rightarrow {\left( {\cfrac{{{d^2}P}}{{d{y^2}}}} \right)_{y = 25}} < 0$

Hence we can say that $P$ attains maximum value at $y = 25$.

Therefore the required numbers are $x = 35 - 25 = 10$ and $y = 25$.