Let x, y 0.

Question

Let x, y > 0. If x3y2 = 215, then the least value of 3x + 2y is

Accepted Answer

x, y > 0 and x3y2 = 215

Now, 3x + 2y = (x + x + x) + (y + y)

So, by A.M $$\ge$$ G.M inequality

$${{3x + 2y} \over 5} \ge \root 5 \of {{x^3}\,.\,{y^2}} $$

$$\therefore$$ $$3x + 2y \ge 5\root 5 \of {{2^{15}}}  \ge 40$$

$$\therefore$$ Least value of $3x + 4y = 40$

Solutions · 1