If $x+2y+3z=10$ find the maximum value of $x^2y^2z^6$. If $x+2y+3z=10$ where x,y,z are postive real numbers, find the maximum value of $x^2y^2z^6$.
I started by trying to find some integer solutions. The first thing I found was $(x,y,z)=(1,3,1)$. But the value of $x^2y^2z^6$ was way too small (it was a multiple choice question in the paper I was solving).
The next thing I tried was setting $x=y=z$ because I thought the product of numbers would be highest if they were equal. And I got $x=y=z=\frac{5}{3}$. And so I got $x^2y^2z^6 = (\frac{5}{3})^{10}$
But this answer seems very weird and even my logic is not right, I merely went by intuition.
Please help me solve this.
 A: I presume $x,y,z$ are non-negative integers. Otherwise, if they are allowed to be negative, we may choose their magnitude to be arbitrarily large.
Write : \begin{align*}
10 &= x + 2y + 3z\\
&= \frac{x}{2} + \frac{x}{2} + y + y + \frac{z}{2} + \frac{z}{2} + \frac{z}{2} + \frac{z}{2} + \frac{z}{2} + \frac{z}{2}\\
&\geqslant 10\cdot\sqrt[10]{\frac{x}{2} \cdot \frac{x}{2} \cdot y \cdot y \cdot \frac{z}{2} \cdot \frac{z}{2} \cdot \frac{z}{2} \cdot \frac{z}{2} \cdot \frac{z}{2} \cdot \frac{z}{2}}\\
&= 10 \cdot \sqrt[10]{\frac{x^2y^2z^6}{2^8}}\\
\end{align*}
We have used simple AM-GM in the first inequality step. Thus, $x^2y^2z^6 \leqslant 2^8 = 256$ , i.e. its maximum possible value is $256$.

The maximum value is attained when : $$\frac{x}{2}=y=\frac{z}{2}\quad\implies\quad x=2y=z\quad\implies\quad (x,y,z) = (2,1,2)$$
A: $10=x+2y+3z=x+2y+z+z+z\geq 5\sqrt[5]{2xyz^3}$
$xyz^3\leq 16\Longrightarrow x^2y^2z^6\leq 256$
Maximum value is at $x=2y=z$ . It's $(2,1,2)$
A: Apply weightedAM-GM to $p,q,r$ with frequencies as 2,2,6, then
$$\frac{2p+2q+6r}{10} \ge (p^2 q^2 r^6)^{1/10}~~~~(1)$$
given that $x+2y+3z=10$ choose $p=x/2, q=y, r=z/2$
Inserting them in (1), we get
$$\frac{10}{10} \ge [(x/2)^2 y^2 (z/2)^6]^{1/10} \implies x^2y^2z^6\le 256.$$
Equality holds when $x/2=y=z/2 \implies x=2y,y=1,z=2y.$
A: A.M. $\geq$ G.M.
$$10=x+2y+3z=\frac{x}{2}+\frac{x}{2}+y+y+\frac{z}{2}+\frac{z}{2}+\frac{z}{2}+\frac{z}{2}+\frac{z}{2}+\frac{z}{2} \geq 10{(\frac{x^2}{4}y^2\frac{z^6}{64})^\frac{1}{10}} $$
$$1\geq (\frac{x^2}{4}y^2\frac{z^6}{64})^\frac{1}{10}$$
$$x^2y^2z^6\leq 2^{10}$$
Equilaty holds when $\frac{x}{2}=y=\frac{z}{2}$
A: Given that you've listed this under precalculus, I rather feel like AM/GM is beyond what you'd be expected to know? Possibly not, it's hard to tell. So let's examine a different method.
My early thoughts were this: Simplify anything we can.
(1) Finding the maximum of $x^2y^2z^6$ is the same as finding the maximum of its square root (assuming the maximum is greater than $1$, which we can easily get by putting in random numbers), so we can instead look at $xyz^3$.
(2) We can convert $xyz^3$ to an expression in two variables by solving the first equation for $x$ and substituting:
$$x = 10 - 2y - 3z \implies xyz^3 = yz^3(10-2y-3z) = 10yz^3 - 2y^2z^3 - 3yz^4$$
We look for the maximum of this function, $f(y,z) = 10yz^3 - 2y^2z^3 - 3yz^4$, and then calculate $x$ at the end.
Normally to find a maximum, we want to find a place where $\frac{d}{dx} f(x) = 0$. But we have two variables here! What are we going to do? We're going to cheat take advantage of partial differentiation. We can differentiate with respect to each variable separately, treating the other temporarily as a constant. Man, it sure feels like cheating. But it works nicely, and even has its own symbol ($\partial$):
$$
\begin{gather*}
\textstyle \frac{\partial}{\partial y} (10yz^3 - 2y^2z^3 - 3yz^4) = 10z^3-4yz^3-3z^4 = 0 \\
10z^3-4yz^3-3z^4 = 0 \implies 10-4y-3z=0 \implies 4y+3z = 10 \\
\textstyle \frac{\partial}{\partial z} (10yz^3 - 2y^2z^3 - 3yz^4) = 30yz^2 - 6y^2z^2 - 12yz^3 = 0 \\
30yz^2 - 6y^2z^2 - 12yz^3 = 0 \implies 30 - 6y - 12z = 0 \implies 6y + 12z = 30
\end{gather*}
$$
Note that we can safely divide by $z^3$ and $yz^2$ respectively, because we know $x, y, z \neq 0$. We now have a linear system $(4y + 3z = 10, 6y + 12z = 30)$ in two variables. Use your favorite solution method, and you'll get $(y, z) = (1, 2)$, and quick calculations with the original equation gives $x = 2$.
Which is the same answer others have given above, with $\max(x^2y^2z^6) = 256$, and no one had to take the 10th root or 5th power of anything.
