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Math Help - show smth has a solution?

  1. #1
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    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
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by cp05 View Post
    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 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 \varphi assumes a maximum on \mathbb{S}^1

    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 \mathbb{S}^1 connected and so \varphi(\mathbb{S}^1)\subseteq\mathbb{R} is an interval.
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  3. #3
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    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!
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by cp05 View Post
    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}
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  5. #5
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    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?
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by cp05 View Post
    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.
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