Exercice XV num 9 - Calculus Made Easy












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Divide π into 3 parts such that the product of their sines may be a maximum or minimum.
What is the most intuitive way to solve it?










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    Do you mean: $$ text{Maximize} qquad sin{A}sin{B}sin{C} qquad text{subject to}~A+B+C=pi $$ ?
    – Matti P.
    Jan 4 at 10:42










  • @MattiP. I guess you'd also need $A,B,C geq 0,$
    – Patricio
    Jan 4 at 10:46










  • @MattiP. I quoted the exercise. Your assumption looks about right.
    – LeoBonhart
    Jan 4 at 10:50






  • 1




    Actually, it should probably be $A, B, C > 0$ since equality would allow one of them to equal $pi$ and the others to equal $0$: an obvious minimum for $sin Acdotsin Bcdotsin C$.
    – KM101
    Jan 4 at 10:50


















0














Divide π into 3 parts such that the product of their sines may be a maximum or minimum.
What is the most intuitive way to solve it?










share|cite|improve this question







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LeoBonhart is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • 1




    Do you mean: $$ text{Maximize} qquad sin{A}sin{B}sin{C} qquad text{subject to}~A+B+C=pi $$ ?
    – Matti P.
    Jan 4 at 10:42










  • @MattiP. I guess you'd also need $A,B,C geq 0,$
    – Patricio
    Jan 4 at 10:46










  • @MattiP. I quoted the exercise. Your assumption looks about right.
    – LeoBonhart
    Jan 4 at 10:50






  • 1




    Actually, it should probably be $A, B, C > 0$ since equality would allow one of them to equal $pi$ and the others to equal $0$: an obvious minimum for $sin Acdotsin Bcdotsin C$.
    – KM101
    Jan 4 at 10:50
















0












0








0







Divide π into 3 parts such that the product of their sines may be a maximum or minimum.
What is the most intuitive way to solve it?










share|cite|improve this question







New contributor




LeoBonhart is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











Divide π into 3 parts such that the product of their sines may be a maximum or minimum.
What is the most intuitive way to solve it?







calculus optimization partial-derivative






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asked Jan 4 at 10:35









LeoBonhartLeoBonhart

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  • 1




    Do you mean: $$ text{Maximize} qquad sin{A}sin{B}sin{C} qquad text{subject to}~A+B+C=pi $$ ?
    – Matti P.
    Jan 4 at 10:42










  • @MattiP. I guess you'd also need $A,B,C geq 0,$
    – Patricio
    Jan 4 at 10:46










  • @MattiP. I quoted the exercise. Your assumption looks about right.
    – LeoBonhart
    Jan 4 at 10:50






  • 1




    Actually, it should probably be $A, B, C > 0$ since equality would allow one of them to equal $pi$ and the others to equal $0$: an obvious minimum for $sin Acdotsin Bcdotsin C$.
    – KM101
    Jan 4 at 10:50
















  • 1




    Do you mean: $$ text{Maximize} qquad sin{A}sin{B}sin{C} qquad text{subject to}~A+B+C=pi $$ ?
    – Matti P.
    Jan 4 at 10:42










  • @MattiP. I guess you'd also need $A,B,C geq 0,$
    – Patricio
    Jan 4 at 10:46










  • @MattiP. I quoted the exercise. Your assumption looks about right.
    – LeoBonhart
    Jan 4 at 10:50






  • 1




    Actually, it should probably be $A, B, C > 0$ since equality would allow one of them to equal $pi$ and the others to equal $0$: an obvious minimum for $sin Acdotsin Bcdotsin C$.
    – KM101
    Jan 4 at 10:50










1




1




Do you mean: $$ text{Maximize} qquad sin{A}sin{B}sin{C} qquad text{subject to}~A+B+C=pi $$ ?
– Matti P.
Jan 4 at 10:42




Do you mean: $$ text{Maximize} qquad sin{A}sin{B}sin{C} qquad text{subject to}~A+B+C=pi $$ ?
– Matti P.
Jan 4 at 10:42












@MattiP. I guess you'd also need $A,B,C geq 0,$
– Patricio
Jan 4 at 10:46




@MattiP. I guess you'd also need $A,B,C geq 0,$
– Patricio
Jan 4 at 10:46












@MattiP. I quoted the exercise. Your assumption looks about right.
– LeoBonhart
Jan 4 at 10:50




@MattiP. I quoted the exercise. Your assumption looks about right.
– LeoBonhart
Jan 4 at 10:50




1




1




Actually, it should probably be $A, B, C > 0$ since equality would allow one of them to equal $pi$ and the others to equal $0$: an obvious minimum for $sin Acdotsin Bcdotsin C$.
– KM101
Jan 4 at 10:50






Actually, it should probably be $A, B, C > 0$ since equality would allow one of them to equal $pi$ and the others to equal $0$: an obvious minimum for $sin Acdotsin Bcdotsin C$.
– KM101
Jan 4 at 10:50












2 Answers
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Let $f:[0,pi]times[0,pi]tomathbb{R}$, $f(x,y)=sin x sin ysin(pi-x-y)$.
You have $frac{partial f}{partial x} =cos xsin ysin(pi-x-y)-sin x sin y cos(pi-x-y)$ and $frac{partial f}{partial y} =sin xcos ysin(pi-x-y)-sin x sin y cos(pi-x-y)$. Set both equal to zero and you find, after dividing by $cos x $ and $ cos(pi-x-y)$ the first and by $cos y$ and $cos(pi-x-y)$ the second, $tan x=tan y=tan(pi-x-y)$.
Hence $x=y=pi/3$.






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    0














    I think it works best with Lagrange multipliers. Let $f(A,B,C) =sin(A)sin(B)sin(C)$ and let $g(A,B,C) = A+B+C = pi$ be the constraint. The system of equations $nabla f =lambda nabla g$ is:



    $$cos(A)sin(B)sin(C) = lambda$$



    $$sin(A)cos(B)sin(C) = lambda$$



    $$sin(A)sin(B)cos(C) =lambda$$



    Subtract the first two equations and cancel $sin(C)$ to get



    $$cos(A)sin(B) - sin(A)cos(B) = 0$$



    or $$sin(A-B) = 0.$$



    Assuming $0<A,B<pi$ we must have $A=B$.



    Similarly we have $B=C$. So the solution is $3A=pi$ or $A=B=C=pi/3$.






    share|cite|improve this answer





















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      2 Answers
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      2 Answers
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      Let $f:[0,pi]times[0,pi]tomathbb{R}$, $f(x,y)=sin x sin ysin(pi-x-y)$.
      You have $frac{partial f}{partial x} =cos xsin ysin(pi-x-y)-sin x sin y cos(pi-x-y)$ and $frac{partial f}{partial y} =sin xcos ysin(pi-x-y)-sin x sin y cos(pi-x-y)$. Set both equal to zero and you find, after dividing by $cos x $ and $ cos(pi-x-y)$ the first and by $cos y$ and $cos(pi-x-y)$ the second, $tan x=tan y=tan(pi-x-y)$.
      Hence $x=y=pi/3$.






      share|cite|improve this answer


























        0














        Let $f:[0,pi]times[0,pi]tomathbb{R}$, $f(x,y)=sin x sin ysin(pi-x-y)$.
        You have $frac{partial f}{partial x} =cos xsin ysin(pi-x-y)-sin x sin y cos(pi-x-y)$ and $frac{partial f}{partial y} =sin xcos ysin(pi-x-y)-sin x sin y cos(pi-x-y)$. Set both equal to zero and you find, after dividing by $cos x $ and $ cos(pi-x-y)$ the first and by $cos y$ and $cos(pi-x-y)$ the second, $tan x=tan y=tan(pi-x-y)$.
        Hence $x=y=pi/3$.






        share|cite|improve this answer
























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          Let $f:[0,pi]times[0,pi]tomathbb{R}$, $f(x,y)=sin x sin ysin(pi-x-y)$.
          You have $frac{partial f}{partial x} =cos xsin ysin(pi-x-y)-sin x sin y cos(pi-x-y)$ and $frac{partial f}{partial y} =sin xcos ysin(pi-x-y)-sin x sin y cos(pi-x-y)$. Set both equal to zero and you find, after dividing by $cos x $ and $ cos(pi-x-y)$ the first and by $cos y$ and $cos(pi-x-y)$ the second, $tan x=tan y=tan(pi-x-y)$.
          Hence $x=y=pi/3$.






          share|cite|improve this answer












          Let $f:[0,pi]times[0,pi]tomathbb{R}$, $f(x,y)=sin x sin ysin(pi-x-y)$.
          You have $frac{partial f}{partial x} =cos xsin ysin(pi-x-y)-sin x sin y cos(pi-x-y)$ and $frac{partial f}{partial y} =sin xcos ysin(pi-x-y)-sin x sin y cos(pi-x-y)$. Set both equal to zero and you find, after dividing by $cos x $ and $ cos(pi-x-y)$ the first and by $cos y$ and $cos(pi-x-y)$ the second, $tan x=tan y=tan(pi-x-y)$.
          Hence $x=y=pi/3$.







          share|cite|improve this answer












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          share|cite|improve this answer










          answered Jan 4 at 11:18









          tommycauterotommycautero

          366




          366























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              I think it works best with Lagrange multipliers. Let $f(A,B,C) =sin(A)sin(B)sin(C)$ and let $g(A,B,C) = A+B+C = pi$ be the constraint. The system of equations $nabla f =lambda nabla g$ is:



              $$cos(A)sin(B)sin(C) = lambda$$



              $$sin(A)cos(B)sin(C) = lambda$$



              $$sin(A)sin(B)cos(C) =lambda$$



              Subtract the first two equations and cancel $sin(C)$ to get



              $$cos(A)sin(B) - sin(A)cos(B) = 0$$



              or $$sin(A-B) = 0.$$



              Assuming $0<A,B<pi$ we must have $A=B$.



              Similarly we have $B=C$. So the solution is $3A=pi$ or $A=B=C=pi/3$.






              share|cite|improve this answer


























                0














                I think it works best with Lagrange multipliers. Let $f(A,B,C) =sin(A)sin(B)sin(C)$ and let $g(A,B,C) = A+B+C = pi$ be the constraint. The system of equations $nabla f =lambda nabla g$ is:



                $$cos(A)sin(B)sin(C) = lambda$$



                $$sin(A)cos(B)sin(C) = lambda$$



                $$sin(A)sin(B)cos(C) =lambda$$



                Subtract the first two equations and cancel $sin(C)$ to get



                $$cos(A)sin(B) - sin(A)cos(B) = 0$$



                or $$sin(A-B) = 0.$$



                Assuming $0<A,B<pi$ we must have $A=B$.



                Similarly we have $B=C$. So the solution is $3A=pi$ or $A=B=C=pi/3$.






                share|cite|improve this answer
























                  0












                  0








                  0






                  I think it works best with Lagrange multipliers. Let $f(A,B,C) =sin(A)sin(B)sin(C)$ and let $g(A,B,C) = A+B+C = pi$ be the constraint. The system of equations $nabla f =lambda nabla g$ is:



                  $$cos(A)sin(B)sin(C) = lambda$$



                  $$sin(A)cos(B)sin(C) = lambda$$



                  $$sin(A)sin(B)cos(C) =lambda$$



                  Subtract the first two equations and cancel $sin(C)$ to get



                  $$cos(A)sin(B) - sin(A)cos(B) = 0$$



                  or $$sin(A-B) = 0.$$



                  Assuming $0<A,B<pi$ we must have $A=B$.



                  Similarly we have $B=C$. So the solution is $3A=pi$ or $A=B=C=pi/3$.






                  share|cite|improve this answer












                  I think it works best with Lagrange multipliers. Let $f(A,B,C) =sin(A)sin(B)sin(C)$ and let $g(A,B,C) = A+B+C = pi$ be the constraint. The system of equations $nabla f =lambda nabla g$ is:



                  $$cos(A)sin(B)sin(C) = lambda$$



                  $$sin(A)cos(B)sin(C) = lambda$$



                  $$sin(A)sin(B)cos(C) =lambda$$



                  Subtract the first two equations and cancel $sin(C)$ to get



                  $$cos(A)sin(B) - sin(A)cos(B) = 0$$



                  or $$sin(A-B) = 0.$$



                  Assuming $0<A,B<pi$ we must have $A=B$.



                  Similarly we have $B=C$. So the solution is $3A=pi$ or $A=B=C=pi/3$.







                  share|cite|improve this answer












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                  share|cite|improve this answer










                  answered Jan 4 at 12:17









                  B. GoddardB. Goddard

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