Absolute maximum for $f(x,y) = 4x^2 + 5y$












0












$begingroup$


Suppose you have a region $R$ that satisfies $4x^2+y^2leq 4$ What is the absolute maximum of the function $f(x,y) = 4x^2 + 5y$ on $R$?



The correct answer is $10$



First, I calculated the critical points in $R$:



$$f_x = 8x$$
$$8x= 0$$
$$x=0$$



$$f_y = 5$$
$$5 ne 0$$



Given the fact that there are no points where both $f_x$ and $f_y$ equal $0$, doesn't that mean there are no critical points in $R$, indicating no maximum value? I don't know if I have to check the boundary in this case, since the region is circular? I am not allowed to use Lagrange Multipliers.



Any insight into how to solve these types of problems would help.






multivariable-calculus optimization







share|cite|improve this question











$endgroup$












  • $begingroup$
    On the boundary substitute $x^2$ from the equation of the boundary into $f(x,y)$ and find and minimise over $y$.
    $endgroup$
    – user121049
    Dec 10 '18 at 8:41
















0












$begingroup$


Suppose you have a region $R$ that satisfies $4x^2+y^2leq 4$ What is the absolute maximum of the function $f(x,y) = 4x^2 + 5y$ on $R$?



The correct answer is $10$



First, I calculated the critical points in $R$:



$$f_x = 8x$$
$$8x= 0$$
$$x=0$$



$$f_y = 5$$
$$5 ne 0$$



Given the fact that there are no points where both $f_x$ and $f_y$ equal $0$, doesn't that mean there are no critical points in $R$, indicating no maximum value? I don't know if I have to check the boundary in this case, since the region is circular? I am not allowed to use Lagrange Multipliers.



Any insight into how to solve these types of problems would help.






multivariable-calculus optimization







share|cite|improve this question











$endgroup$












  • $begingroup$
    On the boundary substitute $x^2$ from the equation of the boundary into $f(x,y)$ and find and minimise over $y$.
    $endgroup$
    – user121049
    Dec 10 '18 at 8:41














0












0








0


3



$begingroup$


Suppose you have a region $R$ that satisfies $4x^2+y^2leq 4$ What is the absolute maximum of the function $f(x,y) = 4x^2 + 5y$ on $R$?



The correct answer is $10$



First, I calculated the critical points in $R$:



$$f_x = 8x$$
$$8x= 0$$
$$x=0$$



$$f_y = 5$$
$$5 ne 0$$



Given the fact that there are no points where both $f_x$ and $f_y$ equal $0$, doesn't that mean there are no critical points in $R$, indicating no maximum value? I don't know if I have to check the boundary in this case, since the region is circular? I am not allowed to use Lagrange Multipliers.



Any insight into how to solve these types of problems would help.






multivariable-calculus optimization







share|cite|improve this question











$endgroup$




Suppose you have a region $R$ that satisfies $4x^2+y^2leq 4$ What is the absolute maximum of the function $f(x,y) = 4x^2 + 5y$ on $R$?



The correct answer is $10$



First, I calculated the critical points in $R$:



$$f_x = 8x$$
$$8x= 0$$
$$x=0$$



$$f_y = 5$$
$$5 ne 0$$



Given the fact that there are no points where both $f_x$ and $f_y$ equal $0$, doesn't that mean there are no critical points in $R$, indicating no maximum value? I don't know if I have to check the boundary in this case, since the region is circular? I am not allowed to use Lagrange Multipliers.



Any insight into how to solve these types of problems would help.






multivariable-calculus optimization




multivariable-calculus optimization






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 8:46







Art

















asked Dec 10 '18 at 8:36









ArtArt

31918




31918












  • $begingroup$
    On the boundary substitute $x^2$ from the equation of the boundary into $f(x,y)$ and find and minimise over $y$.
    $endgroup$
    – user121049
    Dec 10 '18 at 8:41


















  • $begingroup$
    On the boundary substitute $x^2$ from the equation of the boundary into $f(x,y)$ and find and minimise over $y$.
    $endgroup$
    – user121049
    Dec 10 '18 at 8:41
















$begingroup$
On the boundary substitute $x^2$ from the equation of the boundary into $f(x,y)$ and find and minimise over $y$.
$endgroup$
– user121049
Dec 10 '18 at 8:41




$begingroup$
On the boundary substitute $x^2$ from the equation of the boundary into $f(x,y)$ and find and minimise over $y$.
$endgroup$
– user121049
Dec 10 '18 at 8:41










4 Answers
4






active

oldest

votes


















2












$begingroup$

The function $f(x)=x$ also has an absolute maximum on $[-1,1]$, even though there is no point at which $f'(x)=0$.



Remember, the zeroes of the partial derivatives only give you candidates for the critical points in the interior of the domain of $f$. On the border (where a maximum can also be reached sometimes), you can use Lagrange multipliers to calculate the max.



Alternatively, your particular function is fairly simple, so you can do without Lagrange. You have, on the border, $4x^2+y^2=4$, which means you can express $x^2$ as a funciton of $y$, and insert that into $f(x,y)$ to get a function of only $y$, and you can calculate the maximum of $f$ on the border quite easily.






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    When the function is defined by a simple expression ad-hoc methods are recommended. You can find the maximum without using any theorems. $f(x,y) leq 5y+4-y^{2}$ and $y$ is subject to the condition $-2leq y leq 2$. This gives $f(x,y)leq 10$ and this value is attained when $x=0,y=2$.






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      You proved that there are no critical points in the interior of $R$. In order to see what happens at the boundary of $R$, you can use the method of Lagrange multipliers, which leads you to the system:$$left{begin{array}{l}8x=8lambda x\5=8lambda y\4x^2+y^2=4.end{array}right.$$


      Otherwise, you can reduce this to a $1$-dimensional problem: find the maximum of$$(4costheta)^2+5times4sintheta,$$with $thetain[0,2pi]$. It is perhaps better to keep in mind that $cos^2theta+sin^2theta=1$.




      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        I am expected to solve it w/o Lagrange. Is there an alternative?
        $endgroup$
        – Art
        Dec 10 '18 at 8:45






      • 1




        $begingroup$
        I've edited my answer. What do you think now?
        $endgroup$
        – José Carlos Santos
        Dec 10 '18 at 8:51



















      0












      $begingroup$

      We have: $4x^2 + 5y le 4-y^2+5y =10 -y^2+5y-6=10 -(y-2)(y-3)le 10$ since $y le 2$.






      share|cite|improve this answer









      $endgroup$













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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        The function $f(x)=x$ also has an absolute maximum on $[-1,1]$, even though there is no point at which $f'(x)=0$.



        Remember, the zeroes of the partial derivatives only give you candidates for the critical points in the interior of the domain of $f$. On the border (where a maximum can also be reached sometimes), you can use Lagrange multipliers to calculate the max.



        Alternatively, your particular function is fairly simple, so you can do without Lagrange. You have, on the border, $4x^2+y^2=4$, which means you can express $x^2$ as a funciton of $y$, and insert that into $f(x,y)$ to get a function of only $y$, and you can calculate the maximum of $f$ on the border quite easily.






        share|cite|improve this answer











        $endgroup$


















          2












          $begingroup$

          The function $f(x)=x$ also has an absolute maximum on $[-1,1]$, even though there is no point at which $f'(x)=0$.



          Remember, the zeroes of the partial derivatives only give you candidates for the critical points in the interior of the domain of $f$. On the border (where a maximum can also be reached sometimes), you can use Lagrange multipliers to calculate the max.



          Alternatively, your particular function is fairly simple, so you can do without Lagrange. You have, on the border, $4x^2+y^2=4$, which means you can express $x^2$ as a funciton of $y$, and insert that into $f(x,y)$ to get a function of only $y$, and you can calculate the maximum of $f$ on the border quite easily.






          share|cite|improve this answer











          $endgroup$
















            2












            2








            2





            $begingroup$

            The function $f(x)=x$ also has an absolute maximum on $[-1,1]$, even though there is no point at which $f'(x)=0$.



            Remember, the zeroes of the partial derivatives only give you candidates for the critical points in the interior of the domain of $f$. On the border (where a maximum can also be reached sometimes), you can use Lagrange multipliers to calculate the max.



            Alternatively, your particular function is fairly simple, so you can do without Lagrange. You have, on the border, $4x^2+y^2=4$, which means you can express $x^2$ as a funciton of $y$, and insert that into $f(x,y)$ to get a function of only $y$, and you can calculate the maximum of $f$ on the border quite easily.






            share|cite|improve this answer











            $endgroup$



            The function $f(x)=x$ also has an absolute maximum on $[-1,1]$, even though there is no point at which $f'(x)=0$.



            Remember, the zeroes of the partial derivatives only give you candidates for the critical points in the interior of the domain of $f$. On the border (where a maximum can also be reached sometimes), you can use Lagrange multipliers to calculate the max.



            Alternatively, your particular function is fairly simple, so you can do without Lagrange. You have, on the border, $4x^2+y^2=4$, which means you can express $x^2$ as a funciton of $y$, and insert that into $f(x,y)$ to get a function of only $y$, and you can calculate the maximum of $f$ on the border quite easily.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 10 '18 at 8:48

























            answered Dec 10 '18 at 8:40









            5xum5xum

            90.8k394161




            90.8k394161























                2












                $begingroup$

                When the function is defined by a simple expression ad-hoc methods are recommended. You can find the maximum without using any theorems. $f(x,y) leq 5y+4-y^{2}$ and $y$ is subject to the condition $-2leq y leq 2$. This gives $f(x,y)leq 10$ and this value is attained when $x=0,y=2$.






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  When the function is defined by a simple expression ad-hoc methods are recommended. You can find the maximum without using any theorems. $f(x,y) leq 5y+4-y^{2}$ and $y$ is subject to the condition $-2leq y leq 2$. This gives $f(x,y)leq 10$ and this value is attained when $x=0,y=2$.






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    When the function is defined by a simple expression ad-hoc methods are recommended. You can find the maximum without using any theorems. $f(x,y) leq 5y+4-y^{2}$ and $y$ is subject to the condition $-2leq y leq 2$. This gives $f(x,y)leq 10$ and this value is attained when $x=0,y=2$.






                    share|cite|improve this answer









                    $endgroup$



                    When the function is defined by a simple expression ad-hoc methods are recommended. You can find the maximum without using any theorems. $f(x,y) leq 5y+4-y^{2}$ and $y$ is subject to the condition $-2leq y leq 2$. This gives $f(x,y)leq 10$ and this value is attained when $x=0,y=2$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 10 '18 at 8:46









                    Kavi Rama MurthyKavi Rama Murthy

                    60.5k42161




                    60.5k42161























                        1












                        $begingroup$

                        You proved that there are no critical points in the interior of $R$. In order to see what happens at the boundary of $R$, you can use the method of Lagrange multipliers, which leads you to the system:$$left{begin{array}{l}8x=8lambda x\5=8lambda y\4x^2+y^2=4.end{array}right.$$


                        Otherwise, you can reduce this to a $1$-dimensional problem: find the maximum of$$(4costheta)^2+5times4sintheta,$$with $thetain[0,2pi]$. It is perhaps better to keep in mind that $cos^2theta+sin^2theta=1$.




                        share|cite|improve this answer











                        $endgroup$













                        • $begingroup$
                          I am expected to solve it w/o Lagrange. Is there an alternative?
                          $endgroup$
                          – Art
                          Dec 10 '18 at 8:45






                        • 1




                          $begingroup$
                          I've edited my answer. What do you think now?
                          $endgroup$
                          – José Carlos Santos
                          Dec 10 '18 at 8:51
















                        1












                        $begingroup$

                        You proved that there are no critical points in the interior of $R$. In order to see what happens at the boundary of $R$, you can use the method of Lagrange multipliers, which leads you to the system:$$left{begin{array}{l}8x=8lambda x\5=8lambda y\4x^2+y^2=4.end{array}right.$$


                        Otherwise, you can reduce this to a $1$-dimensional problem: find the maximum of$$(4costheta)^2+5times4sintheta,$$with $thetain[0,2pi]$. It is perhaps better to keep in mind that $cos^2theta+sin^2theta=1$.




                        share|cite|improve this answer











                        $endgroup$













                        • $begingroup$
                          I am expected to solve it w/o Lagrange. Is there an alternative?
                          $endgroup$
                          – Art
                          Dec 10 '18 at 8:45






                        • 1




                          $begingroup$
                          I've edited my answer. What do you think now?
                          $endgroup$
                          – José Carlos Santos
                          Dec 10 '18 at 8:51














                        1












                        1








                        1





                        $begingroup$

                        You proved that there are no critical points in the interior of $R$. In order to see what happens at the boundary of $R$, you can use the method of Lagrange multipliers, which leads you to the system:$$left{begin{array}{l}8x=8lambda x\5=8lambda y\4x^2+y^2=4.end{array}right.$$


                        Otherwise, you can reduce this to a $1$-dimensional problem: find the maximum of$$(4costheta)^2+5times4sintheta,$$with $thetain[0,2pi]$. It is perhaps better to keep in mind that $cos^2theta+sin^2theta=1$.




                        share|cite|improve this answer











                        $endgroup$



                        You proved that there are no critical points in the interior of $R$. In order to see what happens at the boundary of $R$, you can use the method of Lagrange multipliers, which leads you to the system:$$left{begin{array}{l}8x=8lambda x\5=8lambda y\4x^2+y^2=4.end{array}right.$$


                        Otherwise, you can reduce this to a $1$-dimensional problem: find the maximum of$$(4costheta)^2+5times4sintheta,$$with $thetain[0,2pi]$. It is perhaps better to keep in mind that $cos^2theta+sin^2theta=1$.





                        share|cite|improve this answer














                        share|cite|improve this answer



                        share|cite|improve this answer








                        edited Dec 10 '18 at 8:51

























                        answered Dec 10 '18 at 8:42









                        José Carlos SantosJosé Carlos Santos

                        161k22128232




                        161k22128232












                        • $begingroup$
                          I am expected to solve it w/o Lagrange. Is there an alternative?
                          $endgroup$
                          – Art
                          Dec 10 '18 at 8:45






                        • 1




                          $begingroup$
                          I've edited my answer. What do you think now?
                          $endgroup$
                          – José Carlos Santos
                          Dec 10 '18 at 8:51


















                        • $begingroup$
                          I am expected to solve it w/o Lagrange. Is there an alternative?
                          $endgroup$
                          – Art
                          Dec 10 '18 at 8:45






                        • 1




                          $begingroup$
                          I've edited my answer. What do you think now?
                          $endgroup$
                          – José Carlos Santos
                          Dec 10 '18 at 8:51
















                        $begingroup$
                        I am expected to solve it w/o Lagrange. Is there an alternative?
                        $endgroup$
                        – Art
                        Dec 10 '18 at 8:45




                        $begingroup$
                        I am expected to solve it w/o Lagrange. Is there an alternative?
                        $endgroup$
                        – Art
                        Dec 10 '18 at 8:45




                        1




                        1




                        $begingroup$
                        I've edited my answer. What do you think now?
                        $endgroup$
                        – José Carlos Santos
                        Dec 10 '18 at 8:51




                        $begingroup$
                        I've edited my answer. What do you think now?
                        $endgroup$
                        – José Carlos Santos
                        Dec 10 '18 at 8:51











                        0












                        $begingroup$

                        We have: $4x^2 + 5y le 4-y^2+5y =10 -y^2+5y-6=10 -(y-2)(y-3)le 10$ since $y le 2$.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          We have: $4x^2 + 5y le 4-y^2+5y =10 -y^2+5y-6=10 -(y-2)(y-3)le 10$ since $y le 2$.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            We have: $4x^2 + 5y le 4-y^2+5y =10 -y^2+5y-6=10 -(y-2)(y-3)le 10$ since $y le 2$.






                            share|cite|improve this answer









                            $endgroup$



                            We have: $4x^2 + 5y le 4-y^2+5y =10 -y^2+5y-6=10 -(y-2)(y-3)le 10$ since $y le 2$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 10 '18 at 8:50









                            DeepSeaDeepSea

                            71.2k54487




                            71.2k54487






























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