Find the greatest value of $a^2\cdot b^3$ where $a+b=10$ Find the greatest value of $a^2\cdot b^3$ where $a,b$ are positive real numbers satisfying $a+b=10$. Determine the values of $a,b$ for which the greatest value is attained.
At first I tried to find the value by putting $a=10-b$, I found a expression, but I got not way to do anything. 
If we apply A.M-G.M. inequality, I think we found the minimum value or something else. It is a pre-Olympiad level problem. So I need some answers or some good hints.
 A: The Am-GM inequality says that $\{a_n\}$ are given then $\frac {\sum a_n}{n} \geq \sqrt[n]{\prod a_n}$.
Rewrite $a+b = 10$ as $2 \times \frac a2 + 3 \times \frac b3 = 10$. Now, apply AM-GM and see what happens, noting that equality occurs when $\frac a2 = \frac b3$.
Applying AM-GM with the set $\{\frac a2,\frac a2,\frac b3,\frac b3,\frac b3\}$, we get:
$$
\frac{\frac a2+\frac a2+\frac b3+\frac b3+\frac b3}{5} \geq \sqrt[5]{\frac a2\cdot\frac a2\cdot\frac b3\cdot\frac b3\cdot\frac b3}
$$
Simplifying, we get that $\frac{a+b}{5} \geq \sqrt[5]{\frac {a^2b^3}{2^23^3}}$.
Knowing that $a+b=10$ gives us:
$$
2 \geq \sqrt[5]{\frac {a^2b^3}{2^23^3}}
$$
Taking the fifth power of both sides:
$$32 \times 2^23^3 \geq a^2b^3$$
which equates to $a^2b^3 \leq 3456$. This is attained when $3a = 2b$ i.e. $a=4,b=6$.
You can also differentiate, but then I believe this is not allowed in Olympiad exams.
A: We have to maximize $(10-b)^2b^3$.
Expand, differentiate, and get $5 b^2 (b^2 - 16 b + 60)=0$.
Therefore, $b=0, 6, 10$. Now test each of these in turn.
We need $a,b \ge 0$ and so $0 \le b \le 10$.
Since $(10-b)^2b^3 \ge 0$ in this interval, the maximum point must be  $b=6$.
A: With the given condition you have to find maximum of $(10-b)^2b^3$.
If you differentiate it with respect to $b$ and equate it to zero. You will get maxima (an easy check on double differentiation will verify it). It will give you $b=0,6,10$. You can easily go further now.
A: Just to give an answer using Lagange multipliers.
Consider $$F=a^2\ b^3+\lambda(a+b-10)$$ COmpute the partial derivatives and set them equal to zero. So $$F'_a=2ab^3+\lambda=0\tag 1$$ $$F'_b=3a^2b^2+\lambda=0\tag 2$$ $$F'_\lambda=a+b-10=0\tag 3$$ Considering $(1)$ and $(2)$, we then have $$2ab^3=3a^2b^2\implies ab^2(2b-3a)=0$$ Then the possible solutions $a=0$ or $b=0$ (both of them would give $a^2b^3=0$) or $b=\frac {3a}2$. Replacing in $F'_\lambda$ then gives $a=4$ and then $b=6$ and then $a^2b^3=3456$.
If the problem was instead $a+b=x$, the solution would have been $a=\frac{2x}5$, $b=\frac{3x}5$, $a^2b^3=\frac{108 x^5}{3125}$.
