Let x,y,z be positive real numbers such that x+y+z=12 and x^3y^4z^5=(0.1)(600)^3.

Question

Let $x,y,z$ be positive real numbers such that $x+y+z=12$ and $x^3y^4z^5=(0.1)(600)^3$. Then $x^3+y^3+z^3$ is equal to:

Accepted Answer

By AM--GM on the 12 numbers $\underbrace{\tfrac x3,\ldots}_{3},\underbrace{\tfrac y4,\ldots}_4,\underbrace{\tfrac z5,\ldots}_5$:
$$\frac{3\left(\frac x3\right)+4\left(\frac y4\right)+5\left(\frac z5\right)}{12}\ge\left(\frac{x}{3}\right)^{3/12}\left(\frac{y}{4}\right)^{4/12}\left(\frac{z}{5}\right)^{5/12},$$
i.e. $1\ge\left(\dfrac{x^3y^4z^5}{3^3\cdot4^4\cdot5^5}\right)^{1/12}$, so $x^3y^4z^5\le3^3\cdot4^4\cdot5^5=(0.1)(600)^3$. Since equality is given, equality must hold in AM--GM, forcing $\dfrac x3=\dfrac y4=\dfrac z5=k$.

With $x+y+z=12$: $3k+4k+5k=12\Rightarrow k=1$, so $x=3,y=4,z=5$.

$x^3+y^3+z^3=27+64+125=216.$

Solutions · 1