Given $f^2(x)+g^2(x)+h^2(x)\leq9$ and $U(x)=3f(x)+4g(x)+10h(x)$.Find the maximum value of $U(x)$

Question

Given $f^2(x)+g^2(x)+h^2(x)\leq9$ and $U(x)=3f(x)+4g(x)+10h(x)$,where $f(x),g(x)$ and $h(x)$ are continuous $\forall x\in R$.Find the maximum value of $U(x)$.

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I tried using AM GM inequality in this question but i could not succeed.This question is given in the vectors section of my book.May be possible,some vectors concept should be applied here.I am not sure about that.
Please help me.Thanks.

Answer 1

Note that $U(x)=\langle 3,4,10\rangle \cdot\langle f(x),g(x),h(x)\rangle$. Also, recall that for vectors $A,B\in \mathbb{R}^3$, $A\cdot B=||A||\,||B||\cos{\theta}$. This implies that $$U(x)=||\langle 3,4,10\rangle ||\,||\langle f(x),g(x),h(x)\rangle ||\cos{\theta}$$ $$=5\sqrt{5}\cos{\theta}\sqrt{f(x)^2+g(x)^2+h(x)^2}$$The maximum of $\cos{\theta}$ occurs when $\theta=0$. Additionally, from the inequality given, the maximum value of $f(x)^2+g(x)^2+h(x)^2$ is 9. Therefore, $$U(x)\leq 5\sqrt{5}(1)\sqrt{9}=15\sqrt{5}$$This maximum is achieved if there exists an x such that $\langle f(x),g(x),h(x)\rangle $ is a multiple of $\langle 3,4,10\rangle $ and $f(x)^2+g(x)^2+h(x)^2=9$.