Question

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?

Answer 1

`n = 1125`

Explanation:

Let `vec v = {f(x),g(x),h(x)}` and `vec v_0 = {3,4,10}`
calling `u_x = << vec v,vec v_0 >>`, `u_x` will attain maximum value when `vec v = lambda vec v_0`, remembering that `norm(vec v) le 9` and that `u_x` is a linear function, it's maximum will be achieved at the frontier, when `norm vec v = 3`.

Putting all together

`<< vec v,vec v_0 >>/(norm(vec v)norm(vec v_0))=1`
`sqrt(n)/(3 sqrt(3^2+4^2+10^2)) = 1`

Solving for `n` gives `n = 1125`
and `lambda = 3/(5 sqrt(5))`
with `vec v_{max} = {9/(5 sqrt[5]), 12/(5 sqrt[5]), 6/sqrt[5]}`

Answer 2

Alternative

Explanation:

For those who know less

Given
`u_x=3f(x)+4g(x)+10h(x)`

Squaring both sides we get

`u_x^2=9f^2(x)+16g^2(x)+100h^2(x)+2*3*4f(x)g(x)+2*4*10g(x)h(x)+2*10*3h(x)f(x)..........(1)`

Now we know for real values

`a^2+b^2>=2ab` Applying this relation we can have the following relations

• `2*3*4f(x)g(x)<=16f^2(x)+9g^2(x).....(2)`

• `2*4*10g(x)h(x)<=16h^2(x)+100g^2(x).......(3)`

• `2*10*3h(x)f(x)<=100f^2(x)+9h^2(x)......(4)`

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

`u_x^2<=9f^2(x)+16g^2(x)+100h^2(x)+16f^2(x)+9g^2(x).+16h^2(x)+100g^2(x)+100f^2(x)+9h^2(x)`

`u_x^2<=125f^2(x)+125g^2(x)+125h^2(x)`

`u_x^2<=125[f^2(x)+g^2(x)+h^2(x)]`

`u_x^2<=125xx9` `" ""since" [f^2(x)+g^2(x)+h^2(x)<=9]`

`=>u_x^2<=1125`

`:.(u_x)_"max"=sqrt1125`

Again given

`(u_x)_"max"=sqrtn`

Hence `n =1125`