Question

The greatest value of $P=a^2{b}^3{c}^4,$ when $a+b+c$ = $18$ & $a,b,c,>0$, is

• A. $P=4^4{.6}^4{.8}^4$
• B. $P=4^2{.6}^3{.8}^4$
• C. $P=3^2{.6}^3{.8}^4$
• D. $P=4^2{.6}^2{.8}^2$

Answer 1

The correct option is B $P=4^2{.6}^3{.8}^4$

Given: $a+b+c=18$ and $a,b,c>0$

To find the greatest value of $P=a^2{b}^3{c}^4$

Sol: Keeping in view that $a+b+c=18$

$a^2{b}^3{c}^4$ is max.

The sum of factors is =$a+b+c=18$

hence product will be max. when all the factors are equal.

$⟹\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=\frac{a+b+c}{2+3+4}=\frac{18}{9}=2$

$a=4,b=6,c=8$

hence maximum value $4^2{6}^3{8}^4$