Find two numbers, $x,y$ whose sum is $35$ and $x^2y^5$ is maximum. 
Find two numbers, $x,y$ whose sum is $35$ and $x^2y^5$ is maximum.

My answer:
$$x+y=35$$
$x^2y^5$ is maximum
$$y=35-x$$
$$\frac{d}{dx} x^2(35-x)^5$$
Which rule to apply here after? I reached:
$$(35-x)^4(-5x^2+(35-x)2x)=0$$
Either $(35-4x)^4 =0$ or $x^2-7=0$

Note that if $y \leq 0 \Rightarrow x^2y^5 \leq 0, \forall x \in \mathbb{R}$, thus assume $y > 0$, and since the max won't change if $x$ is replaced by $-x$, we can also assume $x > 0$. Then we have a much simpler problem to solve.
We have: $35 = x+y = \dfrac{x}{2}+\dfrac{x}{2}+ \dfrac{y}{5}+\dfrac{y}{5}+\dfrac{y}{5}+\dfrac{y}{5}+\dfrac{y}{5} \geq 7\sqrt[7]{\dfrac{x^2y^5}{2^25^5}}\Rightarrow x^2y^5 \leq 2^25^{12}=\text{max}$. Equality occurs when $\dfrac{x}{2}=\dfrac{y}{5}\Rightarrow x = 10,y=25$. Thus we conclude that $x = 10,y = 25$ will do the job.
Note: This answer is not complete and I want to post it as a hint rather than a full answer since you have one part to do: solve the case $x$ is negative.
