Maximize the following equation under a constraint












1












$begingroup$


Maximize $g(x,y)=x^4+y^4$ on $x^2+y^2=9$.



Our professor had sped through it and I didn't fully understand how they arrived at the answer. Some clarification on it would be greatly appreciated. I have attached the writing below:



enter image description here






multivariable-calculus







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Maximize $g(x,y)=x^4+y^4$ on $x^2+y^2=9$.



    Our professor had sped through it and I didn't fully understand how they arrived at the answer. Some clarification on it would be greatly appreciated. I have attached the writing below:



    enter image description here






    multivariable-calculus







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Maximize $g(x,y)=x^4+y^4$ on $x^2+y^2=9$.



      Our professor had sped through it and I didn't fully understand how they arrived at the answer. Some clarification on it would be greatly appreciated. I have attached the writing below:



      enter image description here






      multivariable-calculus







      share|cite|improve this question











      $endgroup$




      Maximize $g(x,y)=x^4+y^4$ on $x^2+y^2=9$.



      Our professor had sped through it and I didn't fully understand how they arrived at the answer. Some clarification on it would be greatly appreciated. I have attached the writing below:



      enter image description here






      multivariable-calculus




      multivariable-calculus






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 22 at 0:58









      David G. Stork

      11k41432




      11k41432










      asked Jan 21 at 23:25









      Random StudentRandom Student

      443




      443






















          2 Answers
          2






          active

          oldest

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          2












          $begingroup$

          $g(x,y) = x^4 + y^4 = x^4 + underbrace{big(sqrt{9-x^2}big)^4}_{from~constraint} = g(x)$



          $${d g(x) over d x} = 4 x^3-4 x left(9-x^2right) = 4x left( x^2 - left(9-x^2right) right) = 4 x^2 (2 x^2 - 9)$$



          Set this function of a single variable $x$ to equal $0$ (for an extremum) and thus find:



          $$left{{xto 0},left{xto -frac{3}{sqrt{2}}right},left{xto frac{3}{sqrt{2}}right}right}$$



          Then test each individually to see if it is a maximum, minimum or inflection point. (You can also calculate second-order derivatives to solve that problem, or plot the functions...)



          enter image description here



          It is clear that $x=0$ corresponds to a local maximum, while the other solutions correspond to local minima.



          By the way, the handwritten notes say "obviously $(0,0)$ is a minimum," which is false. That point does not obey the constraints and is hence invalid.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I'm sorry, I still can't quite understand, the values that you received are the max on the boundary of the graph or just random points on the graph? I'm quite new at this.
            $endgroup$
            – Random Student
            Jan 21 at 23:51












          • $begingroup$
            This approach misses the local extrema at $y=0$.
            $endgroup$
            – random
            Jan 22 at 12:46












          • $begingroup$
            @random: Huh? When $y=0$, $x=pm 3$.
            $endgroup$
            – David G. Stork
            Jan 22 at 17:33










          • $begingroup$
            From the symmetries in the original problem it follows that the local maximum of $g(x,y)$ at $(0,3)$ implies the existence of the same local maximum at $(3,0)$ and $(-3,0)$. The chosen parameterization of the upper half of the circle does show them by the value of g(x) at $x=-3$ and $x=3$, but the value of $g'(x)$ at those points is misleading.
            $endgroup$
            – random
            Jan 23 at 0:39





















          0












          $begingroup$

          Since we are finding a maximum on a circle centered at the origin it makes sense to convert both equations to polar coordinates:



          begin{eqnarray} g(r,theta)&=&(rcostheta)^4+(rsintheta)^4\
          &=&r^4(cos^4theta+sin^4theta)\
          frac{dg}{dr}&=&4r^3(cos^4theta+sin^4theta)=0\
          frac{dg}{dtheta}&=&-4r^4cos^3thetasintheta+4r^4sin^3thetacostheta\
          &=&-4r^4sinthetacostheta(cos^2theta-sin^2theta)\
          &=&-2r^4(2sinthetacostheta)(cos^2theta-sin^2theta)\
          &=&-2r^4sin(2theta)cos(2theta)\
          &=&-r^4sin(4theta)=0
          end{eqnarray}



          So extrema should occur when $sin(4theta)=0$. And $sin(4theta)=0$ only when $4theta$ is a whole multiple of $pi$. Thus you must check points of the circle corresponding to multiples of $dfrac{pi}{4}$.



          You should find that the minima occur when $theta$ is an odd multiple of $dfrac{pi}{4}$ and the maxima will occur when $theta$ is an even multiple of $dfrac{pi}{4}$ (i.e. a multiple of $dfrac{pi}{2})$.



          $$gleft(3,frac{pi}{4}right)=gleft(3,frac{3pi}{4}right)=gleft(3,frac{5pi}{4}right)=gleft(3,frac{7pi}{4}right)=left(pmfrac{3}{sqrt{2}}right)^4+left(pmfrac{3}{sqrt{2}}right)^4=frac{81}{2}$$



          $$g(3,0)=gleft(3,frac{pi}{2}right)=gleft(3,piright)=gleft(3,frac{3pi}{2}right)=3^4=81$$






          share|cite|improve this answer











          $endgroup$













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            2 Answers
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            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

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            active

            oldest

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            active

            oldest

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            2












            $begingroup$

            $g(x,y) = x^4 + y^4 = x^4 + underbrace{big(sqrt{9-x^2}big)^4}_{from~constraint} = g(x)$



            $${d g(x) over d x} = 4 x^3-4 x left(9-x^2right) = 4x left( x^2 - left(9-x^2right) right) = 4 x^2 (2 x^2 - 9)$$



            Set this function of a single variable $x$ to equal $0$ (for an extremum) and thus find:



            $$left{{xto 0},left{xto -frac{3}{sqrt{2}}right},left{xto frac{3}{sqrt{2}}right}right}$$



            Then test each individually to see if it is a maximum, minimum or inflection point. (You can also calculate second-order derivatives to solve that problem, or plot the functions...)



            enter image description here



            It is clear that $x=0$ corresponds to a local maximum, while the other solutions correspond to local minima.



            By the way, the handwritten notes say "obviously $(0,0)$ is a minimum," which is false. That point does not obey the constraints and is hence invalid.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I'm sorry, I still can't quite understand, the values that you received are the max on the boundary of the graph or just random points on the graph? I'm quite new at this.
              $endgroup$
              – Random Student
              Jan 21 at 23:51












            • $begingroup$
              This approach misses the local extrema at $y=0$.
              $endgroup$
              – random
              Jan 22 at 12:46












            • $begingroup$
              @random: Huh? When $y=0$, $x=pm 3$.
              $endgroup$
              – David G. Stork
              Jan 22 at 17:33










            • $begingroup$
              From the symmetries in the original problem it follows that the local maximum of $g(x,y)$ at $(0,3)$ implies the existence of the same local maximum at $(3,0)$ and $(-3,0)$. The chosen parameterization of the upper half of the circle does show them by the value of g(x) at $x=-3$ and $x=3$, but the value of $g'(x)$ at those points is misleading.
              $endgroup$
              – random
              Jan 23 at 0:39


















            2












            $begingroup$

            $g(x,y) = x^4 + y^4 = x^4 + underbrace{big(sqrt{9-x^2}big)^4}_{from~constraint} = g(x)$



            $${d g(x) over d x} = 4 x^3-4 x left(9-x^2right) = 4x left( x^2 - left(9-x^2right) right) = 4 x^2 (2 x^2 - 9)$$



            Set this function of a single variable $x$ to equal $0$ (for an extremum) and thus find:



            $$left{{xto 0},left{xto -frac{3}{sqrt{2}}right},left{xto frac{3}{sqrt{2}}right}right}$$



            Then test each individually to see if it is a maximum, minimum or inflection point. (You can also calculate second-order derivatives to solve that problem, or plot the functions...)



            enter image description here



            It is clear that $x=0$ corresponds to a local maximum, while the other solutions correspond to local minima.



            By the way, the handwritten notes say "obviously $(0,0)$ is a minimum," which is false. That point does not obey the constraints and is hence invalid.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I'm sorry, I still can't quite understand, the values that you received are the max on the boundary of the graph or just random points on the graph? I'm quite new at this.
              $endgroup$
              – Random Student
              Jan 21 at 23:51












            • $begingroup$
              This approach misses the local extrema at $y=0$.
              $endgroup$
              – random
              Jan 22 at 12:46












            • $begingroup$
              @random: Huh? When $y=0$, $x=pm 3$.
              $endgroup$
              – David G. Stork
              Jan 22 at 17:33










            • $begingroup$
              From the symmetries in the original problem it follows that the local maximum of $g(x,y)$ at $(0,3)$ implies the existence of the same local maximum at $(3,0)$ and $(-3,0)$. The chosen parameterization of the upper half of the circle does show them by the value of g(x) at $x=-3$ and $x=3$, but the value of $g'(x)$ at those points is misleading.
              $endgroup$
              – random
              Jan 23 at 0:39
















            2












            2








            2





            $begingroup$

            $g(x,y) = x^4 + y^4 = x^4 + underbrace{big(sqrt{9-x^2}big)^4}_{from~constraint} = g(x)$



            $${d g(x) over d x} = 4 x^3-4 x left(9-x^2right) = 4x left( x^2 - left(9-x^2right) right) = 4 x^2 (2 x^2 - 9)$$



            Set this function of a single variable $x$ to equal $0$ (for an extremum) and thus find:



            $$left{{xto 0},left{xto -frac{3}{sqrt{2}}right},left{xto frac{3}{sqrt{2}}right}right}$$



            Then test each individually to see if it is a maximum, minimum or inflection point. (You can also calculate second-order derivatives to solve that problem, or plot the functions...)



            enter image description here



            It is clear that $x=0$ corresponds to a local maximum, while the other solutions correspond to local minima.



            By the way, the handwritten notes say "obviously $(0,0)$ is a minimum," which is false. That point does not obey the constraints and is hence invalid.






            share|cite|improve this answer











            $endgroup$



            $g(x,y) = x^4 + y^4 = x^4 + underbrace{big(sqrt{9-x^2}big)^4}_{from~constraint} = g(x)$



            $${d g(x) over d x} = 4 x^3-4 x left(9-x^2right) = 4x left( x^2 - left(9-x^2right) right) = 4 x^2 (2 x^2 - 9)$$



            Set this function of a single variable $x$ to equal $0$ (for an extremum) and thus find:



            $$left{{xto 0},left{xto -frac{3}{sqrt{2}}right},left{xto frac{3}{sqrt{2}}right}right}$$



            Then test each individually to see if it is a maximum, minimum or inflection point. (You can also calculate second-order derivatives to solve that problem, or plot the functions...)



            enter image description here



            It is clear that $x=0$ corresponds to a local maximum, while the other solutions correspond to local minima.



            By the way, the handwritten notes say "obviously $(0,0)$ is a minimum," which is false. That point does not obey the constraints and is hence invalid.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 22 at 1:18

























            answered Jan 21 at 23:32









            David G. StorkDavid G. Stork

            11k41432




            11k41432












            • $begingroup$
              I'm sorry, I still can't quite understand, the values that you received are the max on the boundary of the graph or just random points on the graph? I'm quite new at this.
              $endgroup$
              – Random Student
              Jan 21 at 23:51












            • $begingroup$
              This approach misses the local extrema at $y=0$.
              $endgroup$
              – random
              Jan 22 at 12:46












            • $begingroup$
              @random: Huh? When $y=0$, $x=pm 3$.
              $endgroup$
              – David G. Stork
              Jan 22 at 17:33










            • $begingroup$
              From the symmetries in the original problem it follows that the local maximum of $g(x,y)$ at $(0,3)$ implies the existence of the same local maximum at $(3,0)$ and $(-3,0)$. The chosen parameterization of the upper half of the circle does show them by the value of g(x) at $x=-3$ and $x=3$, but the value of $g'(x)$ at those points is misleading.
              $endgroup$
              – random
              Jan 23 at 0:39




















            • $begingroup$
              I'm sorry, I still can't quite understand, the values that you received are the max on the boundary of the graph or just random points on the graph? I'm quite new at this.
              $endgroup$
              – Random Student
              Jan 21 at 23:51












            • $begingroup$
              This approach misses the local extrema at $y=0$.
              $endgroup$
              – random
              Jan 22 at 12:46












            • $begingroup$
              @random: Huh? When $y=0$, $x=pm 3$.
              $endgroup$
              – David G. Stork
              Jan 22 at 17:33










            • $begingroup$
              From the symmetries in the original problem it follows that the local maximum of $g(x,y)$ at $(0,3)$ implies the existence of the same local maximum at $(3,0)$ and $(-3,0)$. The chosen parameterization of the upper half of the circle does show them by the value of g(x) at $x=-3$ and $x=3$, but the value of $g'(x)$ at those points is misleading.
              $endgroup$
              – random
              Jan 23 at 0:39


















            $begingroup$
            I'm sorry, I still can't quite understand, the values that you received are the max on the boundary of the graph or just random points on the graph? I'm quite new at this.
            $endgroup$
            – Random Student
            Jan 21 at 23:51






            $begingroup$
            I'm sorry, I still can't quite understand, the values that you received are the max on the boundary of the graph or just random points on the graph? I'm quite new at this.
            $endgroup$
            – Random Student
            Jan 21 at 23:51














            $begingroup$
            This approach misses the local extrema at $y=0$.
            $endgroup$
            – random
            Jan 22 at 12:46






            $begingroup$
            This approach misses the local extrema at $y=0$.
            $endgroup$
            – random
            Jan 22 at 12:46














            $begingroup$
            @random: Huh? When $y=0$, $x=pm 3$.
            $endgroup$
            – David G. Stork
            Jan 22 at 17:33




            $begingroup$
            @random: Huh? When $y=0$, $x=pm 3$.
            $endgroup$
            – David G. Stork
            Jan 22 at 17:33












            $begingroup$
            From the symmetries in the original problem it follows that the local maximum of $g(x,y)$ at $(0,3)$ implies the existence of the same local maximum at $(3,0)$ and $(-3,0)$. The chosen parameterization of the upper half of the circle does show them by the value of g(x) at $x=-3$ and $x=3$, but the value of $g'(x)$ at those points is misleading.
            $endgroup$
            – random
            Jan 23 at 0:39






            $begingroup$
            From the symmetries in the original problem it follows that the local maximum of $g(x,y)$ at $(0,3)$ implies the existence of the same local maximum at $(3,0)$ and $(-3,0)$. The chosen parameterization of the upper half of the circle does show them by the value of g(x) at $x=-3$ and $x=3$, but the value of $g'(x)$ at those points is misleading.
            $endgroup$
            – random
            Jan 23 at 0:39













            0












            $begingroup$

            Since we are finding a maximum on a circle centered at the origin it makes sense to convert both equations to polar coordinates:



            begin{eqnarray} g(r,theta)&=&(rcostheta)^4+(rsintheta)^4\
            &=&r^4(cos^4theta+sin^4theta)\
            frac{dg}{dr}&=&4r^3(cos^4theta+sin^4theta)=0\
            frac{dg}{dtheta}&=&-4r^4cos^3thetasintheta+4r^4sin^3thetacostheta\
            &=&-4r^4sinthetacostheta(cos^2theta-sin^2theta)\
            &=&-2r^4(2sinthetacostheta)(cos^2theta-sin^2theta)\
            &=&-2r^4sin(2theta)cos(2theta)\
            &=&-r^4sin(4theta)=0
            end{eqnarray}



            So extrema should occur when $sin(4theta)=0$. And $sin(4theta)=0$ only when $4theta$ is a whole multiple of $pi$. Thus you must check points of the circle corresponding to multiples of $dfrac{pi}{4}$.



            You should find that the minima occur when $theta$ is an odd multiple of $dfrac{pi}{4}$ and the maxima will occur when $theta$ is an even multiple of $dfrac{pi}{4}$ (i.e. a multiple of $dfrac{pi}{2})$.



            $$gleft(3,frac{pi}{4}right)=gleft(3,frac{3pi}{4}right)=gleft(3,frac{5pi}{4}right)=gleft(3,frac{7pi}{4}right)=left(pmfrac{3}{sqrt{2}}right)^4+left(pmfrac{3}{sqrt{2}}right)^4=frac{81}{2}$$



            $$g(3,0)=gleft(3,frac{pi}{2}right)=gleft(3,piright)=gleft(3,frac{3pi}{2}right)=3^4=81$$






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              Since we are finding a maximum on a circle centered at the origin it makes sense to convert both equations to polar coordinates:



              begin{eqnarray} g(r,theta)&=&(rcostheta)^4+(rsintheta)^4\
              &=&r^4(cos^4theta+sin^4theta)\
              frac{dg}{dr}&=&4r^3(cos^4theta+sin^4theta)=0\
              frac{dg}{dtheta}&=&-4r^4cos^3thetasintheta+4r^4sin^3thetacostheta\
              &=&-4r^4sinthetacostheta(cos^2theta-sin^2theta)\
              &=&-2r^4(2sinthetacostheta)(cos^2theta-sin^2theta)\
              &=&-2r^4sin(2theta)cos(2theta)\
              &=&-r^4sin(4theta)=0
              end{eqnarray}



              So extrema should occur when $sin(4theta)=0$. And $sin(4theta)=0$ only when $4theta$ is a whole multiple of $pi$. Thus you must check points of the circle corresponding to multiples of $dfrac{pi}{4}$.



              You should find that the minima occur when $theta$ is an odd multiple of $dfrac{pi}{4}$ and the maxima will occur when $theta$ is an even multiple of $dfrac{pi}{4}$ (i.e. a multiple of $dfrac{pi}{2})$.



              $$gleft(3,frac{pi}{4}right)=gleft(3,frac{3pi}{4}right)=gleft(3,frac{5pi}{4}right)=gleft(3,frac{7pi}{4}right)=left(pmfrac{3}{sqrt{2}}right)^4+left(pmfrac{3}{sqrt{2}}right)^4=frac{81}{2}$$



              $$g(3,0)=gleft(3,frac{pi}{2}right)=gleft(3,piright)=gleft(3,frac{3pi}{2}right)=3^4=81$$






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                Since we are finding a maximum on a circle centered at the origin it makes sense to convert both equations to polar coordinates:



                begin{eqnarray} g(r,theta)&=&(rcostheta)^4+(rsintheta)^4\
                &=&r^4(cos^4theta+sin^4theta)\
                frac{dg}{dr}&=&4r^3(cos^4theta+sin^4theta)=0\
                frac{dg}{dtheta}&=&-4r^4cos^3thetasintheta+4r^4sin^3thetacostheta\
                &=&-4r^4sinthetacostheta(cos^2theta-sin^2theta)\
                &=&-2r^4(2sinthetacostheta)(cos^2theta-sin^2theta)\
                &=&-2r^4sin(2theta)cos(2theta)\
                &=&-r^4sin(4theta)=0
                end{eqnarray}



                So extrema should occur when $sin(4theta)=0$. And $sin(4theta)=0$ only when $4theta$ is a whole multiple of $pi$. Thus you must check points of the circle corresponding to multiples of $dfrac{pi}{4}$.



                You should find that the minima occur when $theta$ is an odd multiple of $dfrac{pi}{4}$ and the maxima will occur when $theta$ is an even multiple of $dfrac{pi}{4}$ (i.e. a multiple of $dfrac{pi}{2})$.



                $$gleft(3,frac{pi}{4}right)=gleft(3,frac{3pi}{4}right)=gleft(3,frac{5pi}{4}right)=gleft(3,frac{7pi}{4}right)=left(pmfrac{3}{sqrt{2}}right)^4+left(pmfrac{3}{sqrt{2}}right)^4=frac{81}{2}$$



                $$g(3,0)=gleft(3,frac{pi}{2}right)=gleft(3,piright)=gleft(3,frac{3pi}{2}right)=3^4=81$$






                share|cite|improve this answer











                $endgroup$



                Since we are finding a maximum on a circle centered at the origin it makes sense to convert both equations to polar coordinates:



                begin{eqnarray} g(r,theta)&=&(rcostheta)^4+(rsintheta)^4\
                &=&r^4(cos^4theta+sin^4theta)\
                frac{dg}{dr}&=&4r^3(cos^4theta+sin^4theta)=0\
                frac{dg}{dtheta}&=&-4r^4cos^3thetasintheta+4r^4sin^3thetacostheta\
                &=&-4r^4sinthetacostheta(cos^2theta-sin^2theta)\
                &=&-2r^4(2sinthetacostheta)(cos^2theta-sin^2theta)\
                &=&-2r^4sin(2theta)cos(2theta)\
                &=&-r^4sin(4theta)=0
                end{eqnarray}



                So extrema should occur when $sin(4theta)=0$. And $sin(4theta)=0$ only when $4theta$ is a whole multiple of $pi$. Thus you must check points of the circle corresponding to multiples of $dfrac{pi}{4}$.



                You should find that the minima occur when $theta$ is an odd multiple of $dfrac{pi}{4}$ and the maxima will occur when $theta$ is an even multiple of $dfrac{pi}{4}$ (i.e. a multiple of $dfrac{pi}{2})$.



                $$gleft(3,frac{pi}{4}right)=gleft(3,frac{3pi}{4}right)=gleft(3,frac{5pi}{4}right)=gleft(3,frac{7pi}{4}right)=left(pmfrac{3}{sqrt{2}}right)^4+left(pmfrac{3}{sqrt{2}}right)^4=frac{81}{2}$$



                $$g(3,0)=gleft(3,frac{pi}{2}right)=gleft(3,piright)=gleft(3,frac{3pi}{2}right)=3^4=81$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 26 at 20:37

























                answered Jan 22 at 0:15









                John Wayland BalesJohn Wayland Bales

                14.5k21238




                14.5k21238






























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