Find minimum of $frac{x^2}{2}+x-lambda x$ when $x geq 1, lambda geq 0$












0












$begingroup$


The question is quite straight-forward, how does one find the minimum function (depending on $lambda$) of



$frac{x^2}{2}+x-lambda x$ when $x geq 1, lambda geq 0$?



I know the answer is $frac{3}{2} - lambda$ when $0 leq lambda leq 2$ and $-frac{(1-lambda)^2}{2}$ when $lambda geq 2$, but how does one compute that?






functions optimization convex-optimization







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  • 2




    $begingroup$
    Derivative and set equal to zero.
    $endgroup$
    – Zachary Selk
    Dec 21 '18 at 3:19
















0












$begingroup$


The question is quite straight-forward, how does one find the minimum function (depending on $lambda$) of



$frac{x^2}{2}+x-lambda x$ when $x geq 1, lambda geq 0$?



I know the answer is $frac{3}{2} - lambda$ when $0 leq lambda leq 2$ and $-frac{(1-lambda)^2}{2}$ when $lambda geq 2$, but how does one compute that?






functions optimization convex-optimization







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Derivative and set equal to zero.
    $endgroup$
    – Zachary Selk
    Dec 21 '18 at 3:19














0












0








0





$begingroup$


The question is quite straight-forward, how does one find the minimum function (depending on $lambda$) of



$frac{x^2}{2}+x-lambda x$ when $x geq 1, lambda geq 0$?



I know the answer is $frac{3}{2} - lambda$ when $0 leq lambda leq 2$ and $-frac{(1-lambda)^2}{2}$ when $lambda geq 2$, but how does one compute that?






functions optimization convex-optimization







share|cite|improve this question









$endgroup$




The question is quite straight-forward, how does one find the minimum function (depending on $lambda$) of



$frac{x^2}{2}+x-lambda x$ when $x geq 1, lambda geq 0$?



I know the answer is $frac{3}{2} - lambda$ when $0 leq lambda leq 2$ and $-frac{(1-lambda)^2}{2}$ when $lambda geq 2$, but how does one compute that?






functions optimization convex-optimization




functions optimization convex-optimization






share|cite|improve this question













share|cite|improve this question











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asked Dec 21 '18 at 3:18









ProShitposterProShitposter

827




827








  • 2




    $begingroup$
    Derivative and set equal to zero.
    $endgroup$
    – Zachary Selk
    Dec 21 '18 at 3:19














  • 2




    $begingroup$
    Derivative and set equal to zero.
    $endgroup$
    – Zachary Selk
    Dec 21 '18 at 3:19








2




2




$begingroup$
Derivative and set equal to zero.
$endgroup$
– Zachary Selk
Dec 21 '18 at 3:19




$begingroup$
Derivative and set equal to zero.
$endgroup$
– Zachary Selk
Dec 21 '18 at 3:19










2 Answers
2






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$begingroup$

Note that
$$
x^2/2 +(1-lambda)x = frac{1}{2}left((x+(1-lambda))^2 - (1-lambda)^2right).
$$
From here, I believe you can derive the answer you want.






share|cite|improve this answer









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    1












    $begingroup$

    I would not give the full detail of the answer. But you can consider the following quadratic form of the equation :

    Define $f(x,lambda) = frac{x^{2}}{2}+x - lambda x$ with constraints $xgeq1$ and $lambdageq0$. Now, arrange the function as follows :
    begin{align*}
    f(x,lambda) &= frac{1}{2}(x^{2} +2(1-lambda)x)\
    &= frac{1}{2}bigg((x-(1-lambda))^{2}-(1-lambda)^{2}bigg)
    end{align*}
    Hence, $f$ attains minimum whenever $x=1-lambda$ (I leave the detail for you to figure out by yourself). The rest of the arguments might not be so difficult to follow on your own.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Note that
      $$
      x^2/2 +(1-lambda)x = frac{1}{2}left((x+(1-lambda))^2 - (1-lambda)^2right).
      $$
      From here, I believe you can derive the answer you want.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Note that
        $$
        x^2/2 +(1-lambda)x = frac{1}{2}left((x+(1-lambda))^2 - (1-lambda)^2right).
        $$
        From here, I believe you can derive the answer you want.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Note that
          $$
          x^2/2 +(1-lambda)x = frac{1}{2}left((x+(1-lambda))^2 - (1-lambda)^2right).
          $$
          From here, I believe you can derive the answer you want.






          share|cite|improve this answer









          $endgroup$



          Note that
          $$
          x^2/2 +(1-lambda)x = frac{1}{2}left((x+(1-lambda))^2 - (1-lambda)^2right).
          $$
          From here, I believe you can derive the answer you want.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 21 '18 at 3:30









          induction601induction601

          1,285314




          1,285314























              1












              $begingroup$

              I would not give the full detail of the answer. But you can consider the following quadratic form of the equation :

              Define $f(x,lambda) = frac{x^{2}}{2}+x - lambda x$ with constraints $xgeq1$ and $lambdageq0$. Now, arrange the function as follows :
              begin{align*}
              f(x,lambda) &= frac{1}{2}(x^{2} +2(1-lambda)x)\
              &= frac{1}{2}bigg((x-(1-lambda))^{2}-(1-lambda)^{2}bigg)
              end{align*}
              Hence, $f$ attains minimum whenever $x=1-lambda$ (I leave the detail for you to figure out by yourself). The rest of the arguments might not be so difficult to follow on your own.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                I would not give the full detail of the answer. But you can consider the following quadratic form of the equation :

                Define $f(x,lambda) = frac{x^{2}}{2}+x - lambda x$ with constraints $xgeq1$ and $lambdageq0$. Now, arrange the function as follows :
                begin{align*}
                f(x,lambda) &= frac{1}{2}(x^{2} +2(1-lambda)x)\
                &= frac{1}{2}bigg((x-(1-lambda))^{2}-(1-lambda)^{2}bigg)
                end{align*}
                Hence, $f$ attains minimum whenever $x=1-lambda$ (I leave the detail for you to figure out by yourself). The rest of the arguments might not be so difficult to follow on your own.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  I would not give the full detail of the answer. But you can consider the following quadratic form of the equation :

                  Define $f(x,lambda) = frac{x^{2}}{2}+x - lambda x$ with constraints $xgeq1$ and $lambdageq0$. Now, arrange the function as follows :
                  begin{align*}
                  f(x,lambda) &= frac{1}{2}(x^{2} +2(1-lambda)x)\
                  &= frac{1}{2}bigg((x-(1-lambda))^{2}-(1-lambda)^{2}bigg)
                  end{align*}
                  Hence, $f$ attains minimum whenever $x=1-lambda$ (I leave the detail for you to figure out by yourself). The rest of the arguments might not be so difficult to follow on your own.






                  share|cite|improve this answer









                  $endgroup$



                  I would not give the full detail of the answer. But you can consider the following quadratic form of the equation :

                  Define $f(x,lambda) = frac{x^{2}}{2}+x - lambda x$ with constraints $xgeq1$ and $lambdageq0$. Now, arrange the function as follows :
                  begin{align*}
                  f(x,lambda) &= frac{1}{2}(x^{2} +2(1-lambda)x)\
                  &= frac{1}{2}bigg((x-(1-lambda))^{2}-(1-lambda)^{2}bigg)
                  end{align*}
                  Hence, $f$ attains minimum whenever $x=1-lambda$ (I leave the detail for you to figure out by yourself). The rest of the arguments might not be so difficult to follow on your own.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 21 '18 at 3:36









                  Evan William ChandraEvan William Chandra

                  626313




                  626313






























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