Question

Given $f^2\left(x\right)+g^2\left(x\right)+h^2\left(x\right)\le 9$ and $U\left(x\right)=3f\left(x\right)+4g\left(x\right)+10h\left(x\right)$,where $f\left(x\right).g\left(x\right)$ and $h\left(x\right)$ are continuous $\forall x\in$R. If maximum value of $U\left(x\right)$ is $\sqrt{N}$. Then find $N$.

Answer 1

Given : $U\left(x\right)=3f\left(x\right)+4g\left(x\right)+10h\left(x\right)$

Squaring both sides we get

$U^2\left(x\right)=9f^2\left(x\right)+16g^2\left(x\right)+100h^2\left(x\right)+24f\left(x\right)g\left(x\right)+80g\left(x\right)h\left(x\right)+60f\left(x\right)h\left(x\right)$

we know that,

$a^2+b^2>2ab$

Hence $2ab\le a^2+b^2$

using the above property we can say

$\mathrm{2.3.4}f\left(x\right).g\left(x\right)\le 16f\left(x\right)+0g^2\left(x\right)...\left[1\right]$

$\mathrm{2.4.10}g\left(x\right).h\left(x\right)\le 16h^2\left(x\right)+100g^2\left(x\right)...\left[2\right]$

$\mathrm{2.10.3.}f\left(x\right).h\left(x\right)\le 100f^2\left(x\right)+9h^2\left(x\right)...\left[3\right]$

Using [1] [2] [3] we can write

$U^2\left(x\right)\le 9f^2\left(x\right)+16g^2\left(x\right)+100h^2\left(x\right)+16f^2+9g^2\left(x\right)+16h^2\left(x\right)+100g^2\left(x\right)$

$+100f^2\left(x\right)+9h^2\left(x\right)$

$\Rightarrow U^2\left(x\right)\le 125f^2\left(x\right)+125g^2\left(x\right)+125h^2\left(x\right)$

$\Rightarrow U^2\left(x\right)\le 125\left(f^2\right(x)+g^2(x)+h^2(x)$

Given that $f^2\left(x\right)+g^2\left(x\right)+h^2\left(x\right)\le 9$

$\therefore U^2\left(x\right)\le 125\times 9$

$\Rightarrow U\left(x\right)\le \sqrt{125\times 9}=\sqrt{1125}$

$\therefore$ Maximum value oof $U\left(x\right)=\sqrt{1125}=\sqrt{N}$

$\therefore N=1125$