Ya it's Algebra

x+y=35 $\Rightarrow \dfrac{x}{2}+\dfrac{x}{2}+\dfrac{y}{5}+\dfrac{y}{5}+\dfrac{y}{5}+\dfrac{y}{5}+\dfrac{y}{5}=35$ Applying AM-GM on these 7 terms we can get the maximum value of $x^2y^5$ but since we are interested only in the equality condition which occurs when $\dfrac{x}{2}=\dfrac{y}{5}$ or $\dfrac{x}{y}=\dfrac{2}{5}$ . Using this and $x+y=35$ , we get $x=10$ and $y=25$ . Hence $xy$ = $\Large 25\times 10= \boxed{250}$

$x+y=35\Rightarrow y=35-x$ Subbing into $x^2\cdot y^5$ , we get $x^2y^5=(35-y)^{2}y^5=y^{7}-70y^6+1225y^5$ Now, let $f(y)=y^{7}-420y^6+1225y^5$ . Then, we have $f^\prime(y)=7y^{6}-420y^5+6225y^4=0$ $y^4(7y^2-420y+6125)=0\Rightarrow 7y^4(y-35)(y-25)=0$ We know that when $y=0$ or $y=35$ then $x^2y^5=0$ . Hence, to get the maximum value, $y=25, x=10$ and the answer is 250.