<s>[INST] 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.
 [/INST] 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.
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