Proving the highest product of two numbers with a fixed sum












0












$begingroup$


How to prove that any two numbers with a fixed sum have the highest product when both equal half of the sum?



Example:



Given sum equals $10$.

Why is then $5times5$ more than $6times4, 7times 3, 8times2$, etc.?






products







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Note that $(5+x) + (5-x) = 10$. Now consider $(5+x)(5-x)$ and remember that the square of a number is always non-negative. Adapt this to the general case.
    $endgroup$
    – JMoravitz
    Jan 15 at 17:35












  • $begingroup$
    As an aside, the observation you are asking to prove is a special case of the AM-GM Inequality.
    $endgroup$
    – JMoravitz
    Jan 15 at 17:36










  • $begingroup$
    As a further aside, this generalises. To maximise the product of $n$ (positive) numbers $x_1, dots, x_n$ subject to the constraint that $x_1 + dots + x_n = c$ for some constant $c$, you want to take all of the $x_i$ to be equal to $frac{c}{n}$
    $endgroup$
    – Adam Higgins
    Jan 15 at 17:39


















0












$begingroup$


How to prove that any two numbers with a fixed sum have the highest product when both equal half of the sum?



Example:



Given sum equals $10$.

Why is then $5times5$ more than $6times4, 7times 3, 8times2$, etc.?






products







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Note that $(5+x) + (5-x) = 10$. Now consider $(5+x)(5-x)$ and remember that the square of a number is always non-negative. Adapt this to the general case.
    $endgroup$
    – JMoravitz
    Jan 15 at 17:35












  • $begingroup$
    As an aside, the observation you are asking to prove is a special case of the AM-GM Inequality.
    $endgroup$
    – JMoravitz
    Jan 15 at 17:36










  • $begingroup$
    As a further aside, this generalises. To maximise the product of $n$ (positive) numbers $x_1, dots, x_n$ subject to the constraint that $x_1 + dots + x_n = c$ for some constant $c$, you want to take all of the $x_i$ to be equal to $frac{c}{n}$
    $endgroup$
    – Adam Higgins
    Jan 15 at 17:39
















0












0








0





$begingroup$


How to prove that any two numbers with a fixed sum have the highest product when both equal half of the sum?



Example:



Given sum equals $10$.

Why is then $5times5$ more than $6times4, 7times 3, 8times2$, etc.?






products







share|cite|improve this question











$endgroup$




How to prove that any two numbers with a fixed sum have the highest product when both equal half of the sum?



Example:



Given sum equals $10$.

Why is then $5times5$ more than $6times4, 7times 3, 8times2$, etc.?






products




products






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 15 at 18:24









David G. Stork

12.2k41836




12.2k41836










asked Jan 15 at 17:32









Xxx DddXxx Ddd

554




554








  • 3




    $begingroup$
    Note that $(5+x) + (5-x) = 10$. Now consider $(5+x)(5-x)$ and remember that the square of a number is always non-negative. Adapt this to the general case.
    $endgroup$
    – JMoravitz
    Jan 15 at 17:35












  • $begingroup$
    As an aside, the observation you are asking to prove is a special case of the AM-GM Inequality.
    $endgroup$
    – JMoravitz
    Jan 15 at 17:36










  • $begingroup$
    As a further aside, this generalises. To maximise the product of $n$ (positive) numbers $x_1, dots, x_n$ subject to the constraint that $x_1 + dots + x_n = c$ for some constant $c$, you want to take all of the $x_i$ to be equal to $frac{c}{n}$
    $endgroup$
    – Adam Higgins
    Jan 15 at 17:39
















  • 3




    $begingroup$
    Note that $(5+x) + (5-x) = 10$. Now consider $(5+x)(5-x)$ and remember that the square of a number is always non-negative. Adapt this to the general case.
    $endgroup$
    – JMoravitz
    Jan 15 at 17:35












  • $begingroup$
    As an aside, the observation you are asking to prove is a special case of the AM-GM Inequality.
    $endgroup$
    – JMoravitz
    Jan 15 at 17:36










  • $begingroup$
    As a further aside, this generalises. To maximise the product of $n$ (positive) numbers $x_1, dots, x_n$ subject to the constraint that $x_1 + dots + x_n = c$ for some constant $c$, you want to take all of the $x_i$ to be equal to $frac{c}{n}$
    $endgroup$
    – Adam Higgins
    Jan 15 at 17:39










3




3




$begingroup$
Note that $(5+x) + (5-x) = 10$. Now consider $(5+x)(5-x)$ and remember that the square of a number is always non-negative. Adapt this to the general case.
$endgroup$
– JMoravitz
Jan 15 at 17:35






$begingroup$
Note that $(5+x) + (5-x) = 10$. Now consider $(5+x)(5-x)$ and remember that the square of a number is always non-negative. Adapt this to the general case.
$endgroup$
– JMoravitz
Jan 15 at 17:35














$begingroup$
As an aside, the observation you are asking to prove is a special case of the AM-GM Inequality.
$endgroup$
– JMoravitz
Jan 15 at 17:36




$begingroup$
As an aside, the observation you are asking to prove is a special case of the AM-GM Inequality.
$endgroup$
– JMoravitz
Jan 15 at 17:36












$begingroup$
As a further aside, this generalises. To maximise the product of $n$ (positive) numbers $x_1, dots, x_n$ subject to the constraint that $x_1 + dots + x_n = c$ for some constant $c$, you want to take all of the $x_i$ to be equal to $frac{c}{n}$
$endgroup$
– Adam Higgins
Jan 15 at 17:39






$begingroup$
As a further aside, this generalises. To maximise the product of $n$ (positive) numbers $x_1, dots, x_n$ subject to the constraint that $x_1 + dots + x_n = c$ for some constant $c$, you want to take all of the $x_i$ to be equal to $frac{c}{n}$
$endgroup$
– Adam Higgins
Jan 15 at 17:39












6 Answers
6






active

oldest

votes


















2












$begingroup$

$4xy=(x+y)^2-(x-y)^2$ so since $x+y$ is fixed the product is maximum when $x=y$ so that $x-y=0$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Mark.In 1 line! Very nice.
    $endgroup$
    – Peter Szilas
    Jan 15 at 18:10



















0












$begingroup$

Hint: Let $$a+b=s$$ then $$ab=a(s-a)=-a^2+sa$$ where $s$ is given.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Suppose $x+y=k $, where $k $ is a fixed real number. We want to maximise $xcdot y =x(k-x)=f(x)$. Differentiation w.r.t. $x$ will give us $f'(x)=k-2x$. Equating $k-2x $ to $0$ gives $x=frac k2 implies y=frac k2$.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      You want two numbers which give a particular sum and you want to maximize their product:



      $$m+n = c tag{1}$$



      $$m(n) = ? tag{2}$$



      You can rewrite the first equation in terms of either $m$ or $n$. For instance, if you choose to express it in terms of $n$, you get $n = c-m$. Plugging this in $(2)$, you get



      $$m(c-m) = -m^2+cm$$



      If you know about quadratics, functions in the form $y = ax^2+bx+c$, you could use the fact that $a$ is negative here, so the quadratic has a maximum, which occurs at $h = -frac{b}{2a}$. Using $a = -1$ and $b = c$, you get $-frac{c}{2(-1)} = frac{c}{2}$, so the maximum product occurs when both $m$ and $n$ are half their sum (and are equal to each other).






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        Sometimes a figure is worth 1000 words.



        Given others answerers are providing hints: Can you see the answer in these (tan) plots of $A = x(n-x)$ for various values of $10 leq n leq 20$?



        enter image description here






        share|cite|improve this answer











        $endgroup$





















          0












          $begingroup$

          $$ left( ;left( frac{a+b}{2} right) - t ; right) ; ; left( ; left( frac{a+b}{2} right) + t ; right) ; = ; left( frac{a+b}{2} right)^2 ; - ; ; t^2 $$






          share|cite|improve this answer









          $endgroup$














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            6 Answers
            6






            active

            oldest

            votes








            6 Answers
            6






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            $4xy=(x+y)^2-(x-y)^2$ so since $x+y$ is fixed the product is maximum when $x=y$ so that $x-y=0$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Mark.In 1 line! Very nice.
              $endgroup$
              – Peter Szilas
              Jan 15 at 18:10
















            2












            $begingroup$

            $4xy=(x+y)^2-(x-y)^2$ so since $x+y$ is fixed the product is maximum when $x=y$ so that $x-y=0$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Mark.In 1 line! Very nice.
              $endgroup$
              – Peter Szilas
              Jan 15 at 18:10














            2












            2








            2





            $begingroup$

            $4xy=(x+y)^2-(x-y)^2$ so since $x+y$ is fixed the product is maximum when $x=y$ so that $x-y=0$






            share|cite|improve this answer









            $endgroup$



            $4xy=(x+y)^2-(x-y)^2$ so since $x+y$ is fixed the product is maximum when $x=y$ so that $x-y=0$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 15 at 17:53









            Mark BennetMark Bennet

            82.1k984182




            82.1k984182












            • $begingroup$
              Mark.In 1 line! Very nice.
              $endgroup$
              – Peter Szilas
              Jan 15 at 18:10


















            • $begingroup$
              Mark.In 1 line! Very nice.
              $endgroup$
              – Peter Szilas
              Jan 15 at 18:10
















            $begingroup$
            Mark.In 1 line! Very nice.
            $endgroup$
            – Peter Szilas
            Jan 15 at 18:10




            $begingroup$
            Mark.In 1 line! Very nice.
            $endgroup$
            – Peter Szilas
            Jan 15 at 18:10











            0












            $begingroup$

            Hint: Let $$a+b=s$$ then $$ab=a(s-a)=-a^2+sa$$ where $s$ is given.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Hint: Let $$a+b=s$$ then $$ab=a(s-a)=-a^2+sa$$ where $s$ is given.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Hint: Let $$a+b=s$$ then $$ab=a(s-a)=-a^2+sa$$ where $s$ is given.






                share|cite|improve this answer









                $endgroup$



                Hint: Let $$a+b=s$$ then $$ab=a(s-a)=-a^2+sa$$ where $s$ is given.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 15 at 17:35









                Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                79.6k42867




                79.6k42867























                    0












                    $begingroup$

                    Suppose $x+y=k $, where $k $ is a fixed real number. We want to maximise $xcdot y =x(k-x)=f(x)$. Differentiation w.r.t. $x$ will give us $f'(x)=k-2x$. Equating $k-2x $ to $0$ gives $x=frac k2 implies y=frac k2$.






                    share|cite|improve this answer









                    $endgroup$


















                      0












                      $begingroup$

                      Suppose $x+y=k $, where $k $ is a fixed real number. We want to maximise $xcdot y =x(k-x)=f(x)$. Differentiation w.r.t. $x$ will give us $f'(x)=k-2x$. Equating $k-2x $ to $0$ gives $x=frac k2 implies y=frac k2$.






                      share|cite|improve this answer









                      $endgroup$
















                        0












                        0








                        0





                        $begingroup$

                        Suppose $x+y=k $, where $k $ is a fixed real number. We want to maximise $xcdot y =x(k-x)=f(x)$. Differentiation w.r.t. $x$ will give us $f'(x)=k-2x$. Equating $k-2x $ to $0$ gives $x=frac k2 implies y=frac k2$.






                        share|cite|improve this answer









                        $endgroup$



                        Suppose $x+y=k $, where $k $ is a fixed real number. We want to maximise $xcdot y =x(k-x)=f(x)$. Differentiation w.r.t. $x$ will give us $f'(x)=k-2x$. Equating $k-2x $ to $0$ gives $x=frac k2 implies y=frac k2$.







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Jan 15 at 17:38









                        Thomas ShelbyThomas Shelby

                        4,8082727




                        4,8082727























                            0












                            $begingroup$

                            You want two numbers which give a particular sum and you want to maximize their product:



                            $$m+n = c tag{1}$$



                            $$m(n) = ? tag{2}$$



                            You can rewrite the first equation in terms of either $m$ or $n$. For instance, if you choose to express it in terms of $n$, you get $n = c-m$. Plugging this in $(2)$, you get



                            $$m(c-m) = -m^2+cm$$



                            If you know about quadratics, functions in the form $y = ax^2+bx+c$, you could use the fact that $a$ is negative here, so the quadratic has a maximum, which occurs at $h = -frac{b}{2a}$. Using $a = -1$ and $b = c$, you get $-frac{c}{2(-1)} = frac{c}{2}$, so the maximum product occurs when both $m$ and $n$ are half their sum (and are equal to each other).






                            share|cite|improve this answer









                            $endgroup$


















                              0












                              $begingroup$

                              You want two numbers which give a particular sum and you want to maximize their product:



                              $$m+n = c tag{1}$$



                              $$m(n) = ? tag{2}$$



                              You can rewrite the first equation in terms of either $m$ or $n$. For instance, if you choose to express it in terms of $n$, you get $n = c-m$. Plugging this in $(2)$, you get



                              $$m(c-m) = -m^2+cm$$



                              If you know about quadratics, functions in the form $y = ax^2+bx+c$, you could use the fact that $a$ is negative here, so the quadratic has a maximum, which occurs at $h = -frac{b}{2a}$. Using $a = -1$ and $b = c$, you get $-frac{c}{2(-1)} = frac{c}{2}$, so the maximum product occurs when both $m$ and $n$ are half their sum (and are equal to each other).






                              share|cite|improve this answer









                              $endgroup$
















                                0












                                0








                                0





                                $begingroup$

                                You want two numbers which give a particular sum and you want to maximize their product:



                                $$m+n = c tag{1}$$



                                $$m(n) = ? tag{2}$$



                                You can rewrite the first equation in terms of either $m$ or $n$. For instance, if you choose to express it in terms of $n$, you get $n = c-m$. Plugging this in $(2)$, you get



                                $$m(c-m) = -m^2+cm$$



                                If you know about quadratics, functions in the form $y = ax^2+bx+c$, you could use the fact that $a$ is negative here, so the quadratic has a maximum, which occurs at $h = -frac{b}{2a}$. Using $a = -1$ and $b = c$, you get $-frac{c}{2(-1)} = frac{c}{2}$, so the maximum product occurs when both $m$ and $n$ are half their sum (and are equal to each other).






                                share|cite|improve this answer









                                $endgroup$



                                You want two numbers which give a particular sum and you want to maximize their product:



                                $$m+n = c tag{1}$$



                                $$m(n) = ? tag{2}$$



                                You can rewrite the first equation in terms of either $m$ or $n$. For instance, if you choose to express it in terms of $n$, you get $n = c-m$. Plugging this in $(2)$, you get



                                $$m(c-m) = -m^2+cm$$



                                If you know about quadratics, functions in the form $y = ax^2+bx+c$, you could use the fact that $a$ is negative here, so the quadratic has a maximum, which occurs at $h = -frac{b}{2a}$. Using $a = -1$ and $b = c$, you get $-frac{c}{2(-1)} = frac{c}{2}$, so the maximum product occurs when both $m$ and $n$ are half their sum (and are equal to each other).







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered Jan 15 at 17:53









                                KM101KM101

                                6,0861525




                                6,0861525























                                    0












                                    $begingroup$

                                    Sometimes a figure is worth 1000 words.



                                    Given others answerers are providing hints: Can you see the answer in these (tan) plots of $A = x(n-x)$ for various values of $10 leq n leq 20$?



                                    enter image description here






                                    share|cite|improve this answer











                                    $endgroup$


















                                      0












                                      $begingroup$

                                      Sometimes a figure is worth 1000 words.



                                      Given others answerers are providing hints: Can you see the answer in these (tan) plots of $A = x(n-x)$ for various values of $10 leq n leq 20$?



                                      enter image description here






                                      share|cite|improve this answer











                                      $endgroup$
















                                        0












                                        0








                                        0





                                        $begingroup$

                                        Sometimes a figure is worth 1000 words.



                                        Given others answerers are providing hints: Can you see the answer in these (tan) plots of $A = x(n-x)$ for various values of $10 leq n leq 20$?



                                        enter image description here






                                        share|cite|improve this answer











                                        $endgroup$



                                        Sometimes a figure is worth 1000 words.



                                        Given others answerers are providing hints: Can you see the answer in these (tan) plots of $A = x(n-x)$ for various values of $10 leq n leq 20$?



                                        enter image description here







                                        share|cite|improve this answer














                                        share|cite|improve this answer



                                        share|cite|improve this answer








                                        edited Jan 15 at 18:14

























                                        answered Jan 15 at 17:44









                                        David G. StorkDavid G. Stork

                                        12.2k41836




                                        12.2k41836























                                            0












                                            $begingroup$

                                            $$ left( ;left( frac{a+b}{2} right) - t ; right) ; ; left( ; left( frac{a+b}{2} right) + t ; right) ; = ; left( frac{a+b}{2} right)^2 ; - ; ; t^2 $$






                                            share|cite|improve this answer









                                            $endgroup$


















                                              0












                                              $begingroup$

                                              $$ left( ;left( frac{a+b}{2} right) - t ; right) ; ; left( ; left( frac{a+b}{2} right) + t ; right) ; = ; left( frac{a+b}{2} right)^2 ; - ; ; t^2 $$






                                              share|cite|improve this answer









                                              $endgroup$
















                                                0












                                                0








                                                0





                                                $begingroup$

                                                $$ left( ;left( frac{a+b}{2} right) - t ; right) ; ; left( ; left( frac{a+b}{2} right) + t ; right) ; = ; left( frac{a+b}{2} right)^2 ; - ; ; t^2 $$






                                                share|cite|improve this answer









                                                $endgroup$



                                                $$ left( ;left( frac{a+b}{2} right) - t ; right) ; ; left( ; left( frac{a+b}{2} right) + t ; right) ; = ; left( frac{a+b}{2} right)^2 ; - ; ; t^2 $$







                                                share|cite|improve this answer












                                                share|cite|improve this answer



                                                share|cite|improve this answer










                                                answered Jan 15 at 18:56









                                                Will JagyWill Jagy

                                                105k5103202




                                                105k5103202






























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                                                    Cabo Verde

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                                                      Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
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                                                    Gyllenstierna

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