If $X+Y = 10 $ ($X$ and $Y $ both are positive) then what is the maximum value of $(X^3)(Y^2)$? I can get to the result by trying different values of X and Y but that is of course time taking. I want to know if there is a better way to get to the result?
 A: You can probably use the Arithmetic geometric mean equality. We have:
$$x+y=10$$
$$\frac{x}{3}+\frac{x}{3}+\frac{x}{3}+\frac{y}{2}+\frac{y}{2}=10$$
Since x and y are both positive,
 $$AM\ge GM$$
$$\frac{10}{5} \ge \left(\frac{x^3y^2}{3^32^2}\right)^\frac{1}{5}$$
$$\frac{x^3y^2}{3^32^2} \le 2^5$$
$$x^3y^2 \le 2^73^3$$
A: If, $y = 10 - x$ then the expression can be written as function of one variable:
$$f(x) = x^3(10-x)^2$$
Now all you have to do is to differentiate this function and find its critical points.
A: $Y = 10 - X$,and $X^3Y^2 = X^3(10-X)^2 = X^3(X^2 - 20X + 100) = X^5 - 20X^4 + 100X^3 = f(X)$. 
$f'(X) = 5X^4 - 80X^3 + 300X^2 = 0 \iff 5X^2(X^2 - 16X + 60) = 5X^2(X-6)(X-10) = 0 \iff X = 0, 6, 10$. Since $0 < X < 10$, $X = 6$, and $Y = 10 - 6 = 4$. From this the max value of $X^3Y^2$ is $6^34^2 = 3,456$.
A: You have $y=x-10$, so you're looking for maximum value of 
$$x^3y^2=x^3(10-x)^2=100x^3+x^5-20x^4$$
When $x \to \infty$ then $100x^3+x^5-20x^4 \to \infty$, so there isn't maximal value of $x^3y^3$.
A: We just have to find the stationary points of
$$ f(x) = x^3(10-x)^2 $$
whose derivative is:
$$ f'(x) = 5x^2(10-x)(6-x) $$
hence the maximum of $x^3 y^2$ when $x+y=10$ is attained at $x=6$.
