Suppose $x, y, z$ are positive real number such that $x + 2y + 3z = 1$. Find the maximum value of $xyz^2$

Question

I'm trying to solve this problem using Lagrange's multiplier method but I'm unable to get the value of lambda and also $z.$ kindly help me out with this problem. I'm getting $x = 2y$

I'm stuck somewhere between the steps

Answer 1

By definition, $\lambda$ is a constant, and we don't even really care what $\lambda$ is. It is always linear and factored from the gradient, so instead of finding $\lambda$, use it to jump from one equation to another.

$\lambda=yz^2$

$\lambda=\frac{xz^2}{2}$

$\lambda=\frac{2xyz}{3}$

Answer 2

Hint: prove that $$xyz^2\le \frac{1}{1152}$$ and the equal sign holds if $$x=\frac{1}{4},y=\frac{1}{8},z=\frac{1}{6}$$

Answer 3

Alt. hint (no calculus):   the following inequality holds by AM-GM, with equality iff $\displaystyle x = 2y = \frac{3}{2}z\,$:

$$ \frac{\sqrt{3}}{\sqrt[4]{2}} \cdot \sqrt[4]{xyz^2} = \sqrt[4]{x \cdot 2y \cdot \frac{3}{2}z \cdot \frac{3}{2}z} \;\le\; \frac{x+2y+3z}{4} = \frac{1}{4} \;\;\iff\;\;xyz^2 \le \frac{2}{3^2 \cdot 4^4} = \frac{1}{1152} $$

Answer 4

You can also convert $xyz^2$ into a function of two variables $f(y,z)=(1-2y-3z)yz^2$ and use this test to conclude that the maxima occurs at $(\frac{1}{4},\frac{1}{8}, \frac{1}{6})$ and is $1152.$ You can ignore the other critical points since all of them have at least one $0$ entry while $x,y$ and $z$ are given positive.