Thread: 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

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

2. same kind of problem: show (x-1)^2 + y^2=1 has a solution on the circle x^2+y^2=1

Similarly, x,y)\mapsto (x-1)^2+y^2-1" alt="\varphi:\mathbb{S}^1\to\mathbb{R}x,y)\mapsto (x-1)^2+y^2-1" /> is continuous and connected and so is an interval.

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!

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!

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?

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 for some . The second needs a little more work. Try it out.