show smth has a solution?

• Apr 21st 2010, 08:29 AM
cp05
show smth has a solution?
1. show that max{(x+y)^2 + y^2: x^2 + y^2=1} has a solution
i think i have to use continuity of f from R^2 -> R and connectedness of R and Intermediate 'Value Thm but dont know how :(

2. same kind of problem: show (x-1)^2 + y^2=1 has a solution on the circle x^2+y^2=1
• Apr 21st 2010, 04:41 PM
Drexel28
Quote:

Originally Posted by cp05
1. show that max{(x+y)^2 + y^2: x^2 + y^2=1} has a solution
i think i have to use continuity of f from R^2 -> R and connectedness of R and Intermediate 'Value Thm but dont know how :(

This follows since $\mathbb{S}^1\subseteq\mathbb{R}^2$ is compact and since the map $\varphi:\mathbb{S}^1\to\mathbb{R}:(x,y)\mapsto (x+y)^2+y^2$ is continuous it follows that $\varphi$ assumes a maximum on $\mathbb{S}^1$

Quote:

2. same kind of problem: show (x-1)^2 + y^2=1 has a solution on the circle x^2+y^2=1
Similarly, $\varphi:\mathbb{S}^1\to\mathbb{R}:(x,y)\mapsto (x-1)^2+y^2-1$ is continuous and $\mathbb{S}^1$ connected and so $\varphi(\mathbb{S}^1)\subseteq\mathbb{R}$ is an interval.
• Apr 21st 2010, 06:27 PM
cp05
where does x^2 + y^2 = 1 come into this? For 2, how is it being an interval showing it is on the circle?

Thanks so much!
• Apr 21st 2010, 07:25 PM
Drexel28
Quote:

Originally Posted by cp05
where does x^2 + y^2 = 1 come into this? For 2, how is it being an interval showing it is on the circle?

Thanks so much!

$\mathbb{S}^1=\left\{(x,y)\in\mathbb{R}:x^2+y^2=1\r ight\}=\text{Unit Circle}$
• Apr 21st 2010, 08:01 PM
cp05
Oh ok that makes sense.

So just to double check, since 1. has a maximum, that maximum is the solution
and since 2. is an interval, everything on that interval is a solution, so we've shown it has at least 1 solution?
• Apr 21st 2010, 08:03 PM
Drexel28
Quote:

Originally Posted by cp05
Oh ok that makes sense.

So just to double check, since 1. has a maximum, that maximum is the solution
and since 2. is an interval, everything on that interval is a solution, so we've shown it has at least 1 solution?

For the first one you may conclude that $\sup\varphi(\mathbb{S}^1)=\varphi(s)$ for some $s\in\mathbb{S}^1$. The second needs a little more work. Try it out.