If""f^2(x)+g^2(x)+h^2(x)<=9f2(x)+g2(x)+h2(x)≤9 and u_x=3f(x)+4g(x)+10h(x)ux=3f(x)+4g(x)+10h(x), again (u_x)_"max"=sqrtn,"where"" "ninN(ux)max=√n,where n∈N then what is the value of n?

2 Answers
Cesareo R.
Jun 1, 2016

n = 1125n=1125

Explanation:

Let vec v = {f(x),g(x),h(x)}→v={f(x),g(x),h(x)} and vec v_0 = {3,4,10}→v0={3,4,10}
calling u_x = << vec v,vec v_0 >>ux=⟨→v,→v0⟩, u_xux will attain maximum value when vec v = lambda vec v_0→v=λ→v0, remembering that norm(vec v) le 9∥∥→v∥∥≤9 and that u_xux is a linear function, it's maximum will be achieved at the frontier, when norm vec v = 3∥∥→v∥∥=3.

Putting all together

<< vec v,vec v_0 >>/(norm(vec v)norm(vec v_0))=1⟨→v,→v0⟩∥∥→v∥∥∥∥→v0∥∥=1
sqrt(n)/(3 sqrt(3^2+4^2+10^2)) = 1√n3√32+42+102=1

Solving for nn gives n = 1125n=1125
and lambda = 3/(5 sqrt(5))λ=35√5
with vec v_{max} = {9/(5 sqrt[5]), 12/(5 sqrt[5]), 6/sqrt[5]}

P dilip_k
Jun 1, 2016

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

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