Question

If $x,y$ are positive and $x^3{y}^2=6$. The least value of is $4x+3y$ is

• A. \N
• B. 5
• C. 10
• D. 50

Answer 1

The correct option is C 10

$x,y$ positive with $x^3{y}^2=6$ we can use AM-GM inequality by suitable choice of corfficients
$4x+3y=3\left(\frac{4}{3}\right)x+2\left(\frac{3}{2}y\right)=(\frac{4}{3}x+\frac{4}{3}x+\frac{4}{3}x+\frac{3}{2}y+\frac{3}{2}y)$
$\ge 5{\left({\left(\frac{4}{3}x\right)}^3{\left(\frac{3}{2}y\right)}^2\right)}^{\frac{1}{5}}$
$=5{(\frac{64}{27}\times \frac{9}{4})}^{\frac{1}{5}}(x^3{y}^2{)}^{\frac{1}{5}}$
$=5{(\frac{16}{3}\cdot 6)}^{\frac{1}{5}}=5\cdot 2=10$
the least value is $10$