MHB Find Max Function g: $\sqrt{10^{2016}}$

  • Thread starter Thread starter evinda
  • Start date Start date
  • Tags Tags
    Function Maximum
evinda
Gold Member
MHB
Messages
3,741
Reaction score
0
Hello! (Wave)

We are given the function $g: \mathbb{R}^{2016} \to \mathbb{R}$ with $g(y)=\sqrt{10 y_1^2 + 10^2 y_2^2+ \dots+ 10^{2016} y_{2016}^2} where y=(y_1, \dots, y_{2016})$.

I have found that the maximum of this function over the set $\{ y=(y_1, \dots, y_{2016}) \in \mathbb{R}^{2016}: \sum_{i=1}^{2016} y_i^2=1\}$ is $\sqrt{10^{2016}}$. Am I right? (Thinking)
 
Physics news on Phys.org
evinda said:
Hello! (Wave)

We are given the function $g: \mathbb{R}^{2016} \to \mathbb{R}$ with $g(y)=\sqrt{10 y_1^2 + 10^2 y_2^2+ \dots+ 10^{2016} y_{2016}^2} where y=(y_1, \dots, y_{2016})$.

I have found that the maximum of this function over the set $\{ y=(y_1, \dots, y_{2016}) \in \mathbb{R}^{2016}: \sum_{i=1}^{2016} y_i^2=1\}$ is $\sqrt{10^{2016}}$. Am I right? (Thinking)

Yep. You're right! (Nod)
 
I like Serena said:
Yep. You're right! (Nod)

Great... Thank you! (Clapping)
 
Back
Top