Discussion

Explanation:

Given, x³y² = 24

We need to calculate minimum value of A = 2x + 3y

Since the power of x is 3 and y is 2 in x³y² = 24, we rewrite write A as $\frac{2x}{3}$+ $\frac{2x}{3}$+ $\frac{2x}{3}$+ $\frac{3y}{2}$ + $\frac{3y}{2}$

We know, AM ≥ GM

⇒ $\frac{{\displaystyle \frac{2x}{3}}+\frac{2x}{3}+\frac{2x}{3}+{\displaystyle \frac{3y}{2}}+\frac{3y}{2}}{5}$ ≥  $\sqrt[5]{\frac{2x}{3}\times \frac{2x}{3}\times \frac{2x}{3}\times \frac{3y}{2}\times \frac{3y}{2}}$

⇒ 2x + 3y ≥ 5 × $\sqrt[5]{\frac{2}{3}\times x^3{y}^2}$

⇒ 2x + 3y ≥ 5 × $\sqrt[5]{\frac{2}{3}\times 48}$

⇒ 2x + 3y ≥ 5 × $\sqrt[5]{32}$

⇒ 2x + 3y ≥ 10

∴ The least possible value of 2x + 3y = 10

Hence, option (a).

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