Calculus
AM-GM / Optimization
Grade Class 12
Question:
For $x>0$, $y>0$ with $x^2y^3=6$, find $\min(3x+4y)$.
Step-by-Step Solution
Key Concept: Use AM-GM by splitting $3x+4y$ into 5 parts matching powers; balance $x^2y^3$ constraint.
$3x+4y = \frac{3x}{2}+\frac{3x}{2}+\frac{4y}{3}+\frac{4y}{3}+\frac{4y}{3}\geq 5\left(\left(\frac{3}{2}\right)^2\left(\frac{4}{3}\right)^3 x^2 y^3\right)^{1/5}=5\left(\frac{9}{4}\cdot\frac{64}{27}\cdot 6\right)^{1/5}=5\cdot(32)^{1/5}=5\cdot 2=10$. Min $=10$.
Correct Answer: 10