Question

Let $2x+3y+4z=9$, $x$, $y$, $z>0$, then the maximum value of ${(1+x)}^2{(2+y)}^3{(4+z)}^4$ is

• A. $2^9$
• B. $4^9$
• C. ${\left(\frac{11}{3}\right)}^9$
• D. ${\left(\frac{11}{6}\right)}^9$

Answer 1

The correct option is C ${\left(\frac{11}{3}\right)}^9$

Let us consider, $f=(1+x{)}^2(2+y{)}^3(4+z{)}^4$ and $g=2x+3y+4z=9$

We know that,

$\Delta f=\lambda \Delta g$

$2\left[\right(1+x\left)\right(2+y{)}^3(4+z{)}^4,3(1+x{)}^2(2+y{)}^2(4+z{)}^4,4(1+x{)}^2(2+y{)}^3(4+z{)}^3]=\lambda [2,3,4]$

$\Rightarrow 2(1+x)(2+y{)}^3(4+z{)}^4=2\lambda$

$3(1+x{)}^2(2+y{)}^2(4+z{)}^4=3\lambda$

$\Rightarrow 4(1+x{)}^2(2+y{)}^3(4+z{)}^3=4\lambda$

$\lambda =(1+x)(2+y{)}^3(4+z{)}^4=(1+x{)}^2(2+y{)}^2(4+z{)}^4$

$=(1+x{)}^2(2+y{)}^3(4+z{)}^3$

Here, $x,y,z>0$

On solving we get, $(2+y)(4+z)=(1+x)(4+z)=(1+x)(2+y)$

$\Rightarrow (2+y)(4+z)=(1+x)(4+z)and(1+x)(4+z)=(1+x)(2+y)$

$\Rightarrow 2+y=1+x,4+z=2+y$

$\Rightarrow x=1+y,z=y-2$

by substituting the x and z value in g we get $2(1+y)+3y+4(y-2)=9$

Finally , $(x,y,z)=(\frac{8}{3},\frac{5}{3},\frac{-1}{3})$

$f(\frac{8}{3},\frac{5}{3},\frac{-1}{3})$

And the maximum value is $(\frac{11}{3}{)}^9$