# If""f^2(x)+g^2(x)+h^2(x)<=9 and u_x=3f(x)+4g(x)+10h(x), again (u_x)_"max"=sqrtn,"where"" "ninN then what is the value of n?

Jun 1, 2016

$n = 1125$

#### Explanation:

Let $\vec{v} = \left\{f \left(x\right) , g \left(x\right) , h \left(x\right)\right\}$ and ${\vec{v}}_{0} = \left\{3 , 4 , 10\right\}$
calling ${u}_{x} = \left\langle\vec{v} , {\vec{v}}_{0}\right\rangle$, ${u}_{x}$ will attain maximum value when $\vec{v} = \lambda {\vec{v}}_{0}$, remembering that $\left\lVert \vec{v} \right\rVert \le 9$ and that ${u}_{x}$ is a linear function, it's maximum will be achieved at the frontier, when $\left\lVert \vec{v} \right\rVert = 3$.

Putting all together

$\frac{\left\langle\vec{v} , {\vec{v}}_{0}\right\rangle}{\left\lVert \vec{v} \right\rVert \left\lVert {\vec{v}}_{0} \right\rVert} = 1$
$\frac{\sqrt{n}}{3 \sqrt{{3}^{2} + {4}^{2} + {10}^{2}}} = 1$

Solving for $n$ gives $n = 1125$
and $\lambda = \frac{3}{5 \sqrt{5}}$
with ${\vec{v}}_{\max} = \left\{\frac{9}{5 \sqrt{5}} , \frac{12}{5 \sqrt{5}} , \frac{6}{\sqrt{5}}\right\}$

Jun 1, 2016

Alternative

#### Explanation:

For those who know less

Given
${u}_{x} = 3 f \left(x\right) + 4 g \left(x\right) + 10 h \left(x\right)$

Squaring both sides we get

${u}_{x}^{2} = 9 {f}^{2} \left(x\right) + 16 {g}^{2} \left(x\right) + 100 {h}^{2} \left(x\right) + 2 \cdot 3 \cdot 4 f \left(x\right) g \left(x\right) + 2 \cdot 4 \cdot 10 g \left(x\right) h \left(x\right) + 2 \cdot 10 \cdot 3 h \left(x\right) f \left(x\right) \ldots \ldots \ldots . \left(1\right)$

Now we know for real values

${a}^{2} + {b}^{2} \ge 2 a b$ Applying this relation we can have the following relations

• $2 \cdot 3 \cdot 4 f \left(x\right) g \left(x\right) \le 16 {f}^{2} \left(x\right) + 9 {g}^{2} \left(x\right) \ldots . . \left(2\right)$

• $2 \cdot 4 \cdot 10 g \left(x\right) h \left(x\right) \le 16 {h}^{2} \left(x\right) + 100 {g}^{2} \left(x\right) \ldots \ldots . \left(3\right)$

• $2 \cdot 10 \cdot 3 h \left(x\right) f \left(x\right) \le 100 {f}^{2} \left(x\right) + 9 {h}^{2} \left(x\right) \ldots \ldots \left(4\right)$

Combining equations (1),(2),(3)& (4) we get

${u}_{x}^{2} \le 9 {f}^{2} \left(x\right) + 16 {g}^{2} \left(x\right) + 100 {h}^{2} \left(x\right) + 16 {f}^{2} \left(x\right) + 9 {g}^{2} \left(x\right) . + 16 {h}^{2} \left(x\right) + 100 {g}^{2} \left(x\right) + 100 {f}^{2} \left(x\right) + 9 {h}^{2} \left(x\right)$

${u}_{x}^{2} \le 125 {f}^{2} \left(x\right) + 125 {g}^{2} \left(x\right) + 125 {h}^{2} \left(x\right)$

${u}_{x}^{2} \le 125 \left[{f}^{2} \left(x\right) + {g}^{2} \left(x\right) + {h}^{2} \left(x\right)\right]$

${u}_{x}^{2} \le 125 \times 9$$\text{ ""since} \left[{f}^{2} \left(x\right) + {g}^{2} \left(x\right) + {h}^{2} \left(x\right) \le 9\right]$

$\implies {u}_{x}^{2} \le 1125$

$\therefore {\left({u}_{x}\right)}_{\text{max}} = \sqrt{1125}$

Again given

${\left({u}_{x}\right)}_{\text{max}} = \sqrt{n}$

Hence $n = 1125$