Question

If $x,y,z$ and $w$ are positive integers such that $x+2y+3z$ $+4w=50,$ then maximum value of ${\left(\frac{x^2{y}^4{z}^{3}w}{16}\right)}^{1/10}$ is

• A. 4
• B. 5
• C. 6
• D. 7

Answer 1

Answer: (B)

We have $x+2y+3z+4w=50$
Using the fact A.M. $\ge$ G.M., we get
$\frac{2\left(\frac{x}{2}\right)+4\left(\frac{y}{2}\right)+3\left(\frac{z}{1}\right)+1\left(\frac{4w}{1}\right)}{2+4+3+1}=\frac{50}{10}\ge {\left[{\left(\frac{x}{2}\right)}^2{\left(\frac{y}{2}\right)}^4(z{)}^3(4w)\right]}^{1/10}$
$\Rightarrow 5\ge {\left[\left(\frac{x^2}{2^2}\right)\left(\frac{y^4}{2^4}\right)(z{)}^3\left(2^{2}w\right)\right]}^{1/10}$
$\Rightarrow 5\ge {\left(\frac{x^2{y}^4{z}^{3}w}{16}\right)}^{1/10}$