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?
calculus optimization partial-derivative
asked Jan 4 at 10:35
LeoBonhartLeoBonhart
134
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add a comment |
$begingroup$
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
asked Jan 4 at 10:35
LeoBonhartLeoBonhart
134
$endgroup$
1
$begingroup$
Do you mean: $$ text{Maximize} qquad sin{A}sin{B}sin{C} qquad text{subject to}~A+B+C=pi $$ ?
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– Matti P.
Jan 4 at 10:42
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@MattiP. I guess you'd also need $A,B,C geq 0,$
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– Patricio
Jan 4 at 10:46
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@MattiP. I quoted the exercise. Your assumption looks about right.
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– LeoBonhart
Jan 4 at 10:50
1
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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 $sin Acdotsin Bcdotsin C$.
$endgroup$
– KM101
Jan 4 at 10:50
add a comment |
$begingroup$
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
asked Jan 4 at 10:35
LeoBonhartLeoBonhart
134
$endgroup$
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
calculus optimization partial-derivative
asked Jan 4 at 10:35
LeoBonhartLeoBonhart
134
asked Jan 4 at 10:35
LeoBonhartLeoBonhart
134
asked Jan 4 at 10:35
LeoBonhartLeoBonhart
134
asked Jan 4 at 10:35
LeoBonhartLeoBonhart
134
asked Jan 4 at 10:35
LeoBonhartLeoBonhart
134
134
1
$begingroup$
Do you mean: $$ text{Maximize} qquad sin{A}sin{B}sin{C} qquad text{subject to}~A+B+C=pi $$ ?
$endgroup$
– Matti P.
Jan 4 at 10:42
$begingroup$
@MattiP. I guess you'd also need $A,B,C geq 0,$
$endgroup$
– Patricio
Jan 4 at 10:46
$begingroup$
@MattiP. I quoted the exercise. Your assumption looks about right.
$endgroup$
– LeoBonhart
Jan 4 at 10:50
1
$begingroup$
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 $sin Acdotsin Bcdotsin C$.
$endgroup$
– KM101
Jan 4 at 10:50
add a comment |
1
$begingroup$
Do you mean: $$ text{Maximize} qquad sin{A}sin{B}sin{C} qquad text{subject to}~A+B+C=pi $$ ?
$endgroup$
– Matti P.
Jan 4 at 10:42
$begingroup$
@MattiP. I guess you'd also need $A,B,C geq 0,$
$endgroup$
– Patricio
Jan 4 at 10:46
$begingroup$
@MattiP. I quoted the exercise. Your assumption looks about right.
$endgroup$
– LeoBonhart
Jan 4 at 10:50
1
$begingroup$
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 $sin Acdotsin Bcdotsin C$.
$endgroup$
– KM101
Jan 4 at 10:50
1
1
$begingroup$
Do you mean: $$ text{Maximize} qquad sin{A}sin{B}sin{C} qquad text{subject to}~A+B+C=pi $$ ?
$endgroup$
– Matti P.
Jan 4 at 10:42
$begingroup$
Do you mean: $$ text{Maximize} qquad sin{A}sin{B}sin{C} qquad text{subject to}~A+B+C=pi $$ ?
$endgroup$
– Matti P.
Jan 4 at 10:42
$begingroup$
@MattiP. I guess you'd also need $A,B,C geq 0,$
$endgroup$
– Patricio
Jan 4 at 10:46
$begingroup$
@MattiP. I guess you'd also need $A,B,C geq 0,$
$endgroup$
– Patricio
Jan 4 at 10:46
$begingroup$
@MattiP. I quoted the exercise. Your assumption looks about right.
$endgroup$
– LeoBonhart
Jan 4 at 10:50
$begingroup$
@MattiP. I quoted the exercise. Your assumption looks about right.
$endgroup$
– LeoBonhart
Jan 4 at 10:50
1
1
$begingroup$
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 $sin Acdotsin Bcdotsin C$.
$endgroup$
– KM101
Jan 4 at 10:50
$begingroup$
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 $sin Acdotsin Bcdotsin C$.
$endgroup$
– KM101
Jan 4 at 10:50
add a comment |
2 Answers
2
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oldest
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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 $x=y=pi/3$.
answered Jan 4 at 11:18
tommycauterotommycautero
888
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add a comment |
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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 $A=B=C=pi/3$.
answered Jan 4 at 12:17
B. GoddardB. Goddard
19.4k21442
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2 Answers
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$begingroup$
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 $x=y=pi/3$.
answered Jan 4 at 11:18
tommycauterotommycautero
888
$endgroup$
add a comment |
$begingroup$
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 $x=y=pi/3$.
answered Jan 4 at 11:18
tommycauterotommycautero
888
$endgroup$
add a comment |
$begingroup$
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 $x=y=pi/3$.
answered Jan 4 at 11:18
tommycauterotommycautero
888
$endgroup$
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 $x=y=pi/3$.
answered Jan 4 at 11:18
tommycauterotommycautero
888
answered Jan 4 at 11:18
tommycauterotommycautero
888
answered Jan 4 at 11:18
tommycauterotommycautero
888
answered Jan 4 at 11:18
tommycauterotommycautero
888
888
add a comment |
add a comment |
$begingroup$
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 $A=B=C=pi/3$.
answered Jan 4 at 12:17
B. GoddardB. Goddard
19.4k21442
$endgroup$
add a comment |
$begingroup$
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 $A=B=C=pi/3$.
answered Jan 4 at 12:17
B. GoddardB. Goddard
19.4k21442
$endgroup$
add a comment |
$begingroup$
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 $A=B=C=pi/3$.
answered Jan 4 at 12:17
B. GoddardB. Goddard
19.4k21442
$endgroup$
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 $A=B=C=pi/3$.
answered Jan 4 at 12:17
B. GoddardB. Goddard
19.4k21442
answered Jan 4 at 12:17
B. GoddardB. Goddard
19.4k21442
answered Jan 4 at 12:17
B. GoddardB. Goddard
19.4k21442
answered Jan 4 at 12:17
B. GoddardB. Goddard
19.4k21442
19.4k21442
add a comment |
add a comment |
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1
$begingroup$
Do you mean: $$ text{Maximize} qquad sin{A}sin{B}sin{C} qquad text{subject to}~A+B+C=pi $$ ?
$endgroup$
– Matti P.
Jan 4 at 10:42
$begingroup$
@MattiP. I guess you'd also need $A,B,C geq 0,$
$endgroup$
– Patricio
Jan 4 at 10:46
$begingroup$
@MattiP. I quoted the exercise. Your assumption looks about right.
$endgroup$
– LeoBonhart
Jan 4 at 10:50
1
$begingroup$
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 $sin Acdotsin Bcdotsin C$.
$endgroup$
– KM101
Jan 4 at 10:50