Question

Let $x,y>0$. If $x^3{y}^2=2^{15}$, then the least value of $3x+2y$ is

• A. 30
• B. 32
• C. 36
• D. 40

Answer 1

Answer: (D)

Using $AM\ge GM$
$\frac{x+x+x+y+y}{5}\ge {\left(x^3\cdot y^2\right)}^{\frac{1}{5}}$
$\frac{3x+2y}{5}\ge {\left(2^{15}\right)}^{\frac{1}{5}}$
$(3x+2y{)}_{\min }=40$