Finding the maximum of a function on a triangle
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I want to find the maximum of $f(x,y) = x^ae^{-x}y^be^{-y}$ on the triangle given by $xgeq0$, $ygeq0$, and $x+yleq1$ in terms of $a$ and $b$ such that $a,b>0$.
I can see that the vertices of the triangle are $(0,0)$, $(1,0)$, and $(0,1)$. To check whether there is a maximum in the interior of the triangle, would I just check to find points where both partials vanish, or do I need to apply Lagrange multipliers? If so, what would I use for the constraint function? On the boundaries, I know to check the values at the three vertices, but what about all the other points on the boundaries?
multivariable-calculus lagrange-multiplier
asked Dec 2 at 21:26
chris102212
444
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up vote
2
down vote
favorite
I want to find the maximum of $f(x,y) = x^ae^{-x}y^be^{-y}$ on the triangle given by $xgeq0$, $ygeq0$, and $x+yleq1$ in terms of $a$ and $b$ such that $a,b>0$.
I can see that the vertices of the triangle are $(0,0)$, $(1,0)$, and $(0,1)$. To check whether there is a maximum in the interior of the triangle, would I just check to find points where both partials vanish, or do I need to apply Lagrange multipliers? If so, what would I use for the constraint function? On the boundaries, I know to check the values at the three vertices, but what about all the other points on the boundaries?
multivariable-calculus lagrange-multiplier
asked Dec 2 at 21:26
chris102212
444
Lagrange multipliers are only used when you are forced to a constraint in the form of an equality; i.e. "maximise/minimise $f(x)$ given $g(x)=0$"" type questions. You would almost never want to try to introduce $g$ when it's not already there.
– YiFan
Dec 2 at 21:37
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I want to find the maximum of $f(x,y) = x^ae^{-x}y^be^{-y}$ on the triangle given by $xgeq0$, $ygeq0$, and $x+yleq1$ in terms of $a$ and $b$ such that $a,b>0$.
I can see that the vertices of the triangle are $(0,0)$, $(1,0)$, and $(0,1)$. To check whether there is a maximum in the interior of the triangle, would I just check to find points where both partials vanish, or do I need to apply Lagrange multipliers? If so, what would I use for the constraint function? On the boundaries, I know to check the values at the three vertices, but what about all the other points on the boundaries?
multivariable-calculus lagrange-multiplier
asked Dec 2 at 21:26
chris102212
444
I want to find the maximum of $f(x,y) = x^ae^{-x}y^be^{-y}$ on the triangle given by $xgeq0$, $ygeq0$, and $x+yleq1$ in terms of $a$ and $b$ such that $a,b>0$.
I can see that the vertices of the triangle are $(0,0)$, $(1,0)$, and $(0,1)$. To check whether there is a maximum in the interior of the triangle, would I just check to find points where both partials vanish, or do I need to apply Lagrange multipliers? If so, what would I use for the constraint function? On the boundaries, I know to check the values at the three vertices, but what about all the other points on the boundaries?
multivariable-calculus lagrange-multiplier
multivariable-calculus lagrange-multiplier
asked Dec 2 at 21:26
chris102212
444
asked Dec 2 at 21:26
chris102212
444
asked Dec 2 at 21:26
chris102212
444
asked Dec 2 at 21:26
chris102212
444
asked Dec 2 at 21:26
chris102212
444
444
Lagrange multipliers are only used when you are forced to a constraint in the form of an equality; i.e. "maximise/minimise $f(x)$ given $g(x)=0$"" type questions. You would almost never want to try to introduce $g$ when it's not already there.
– YiFan
Dec 2 at 21:37
add a comment |
Lagrange multipliers are only used when you are forced to a constraint in the form of an equality; i.e. "maximise/minimise $f(x)$ given $g(x)=0$"" type questions. You would almost never want to try to introduce $g$ when it's not already there.
– YiFan
Dec 2 at 21:37
Lagrange multipliers are only used when you are forced to a constraint in the form of an equality; i.e. "maximise/minimise $f(x)$ given $g(x)=0$"" type questions. You would almost never want to try to introduce $g$ when it's not already there.
– YiFan
Dec 2 at 21:37
Lagrange multipliers are only used when you are forced to a constraint in the form of an equality; i.e. "maximise/minimise $f(x)$ given $g(x)=0$"" type questions. You would almost never want to try to introduce $g$ when it's not already there.
– YiFan
Dec 2 at 21:37
add a comment |
1 Answer
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On two of the three boundaries, the function is just zero. On the last one, you know $x + y = 1$, so you can rewrite your function as a one-variable function
$$f(x, y) = f(x, 1 - x) = x^a e^{-x} (1 - x)^b e^{-(1 - x)} = frac 1 e x^a (1 - x)^b$$
which isn't too hard to maximize.
Then for the interior, try to identify the critical points where $f_x = f_y = 0$.
answered Dec 2 at 21:28
T. Bongers
22.5k54460
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
On two of the three boundaries, the function is just zero. On the last one, you know $x + y = 1$, so you can rewrite your function as a one-variable function
$$f(x, y) = f(x, 1 - x) = x^a e^{-x} (1 - x)^b e^{-(1 - x)} = frac 1 e x^a (1 - x)^b$$
which isn't too hard to maximize.
Then for the interior, try to identify the critical points where $f_x = f_y = 0$.
answered Dec 2 at 21:28
T. Bongers
22.5k54460
add a comment |
up vote
1
down vote
accepted
On two of the three boundaries, the function is just zero. On the last one, you know $x + y = 1$, so you can rewrite your function as a one-variable function
$$f(x, y) = f(x, 1 - x) = x^a e^{-x} (1 - x)^b e^{-(1 - x)} = frac 1 e x^a (1 - x)^b$$
which isn't too hard to maximize.
Then for the interior, try to identify the critical points where $f_x = f_y = 0$.
answered Dec 2 at 21:28
T. Bongers
22.5k54460
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
On two of the three boundaries, the function is just zero. On the last one, you know $x + y = 1$, so you can rewrite your function as a one-variable function
$$f(x, y) = f(x, 1 - x) = x^a e^{-x} (1 - x)^b e^{-(1 - x)} = frac 1 e x^a (1 - x)^b$$
which isn't too hard to maximize.
Then for the interior, try to identify the critical points where $f_x = f_y = 0$.
answered Dec 2 at 21:28
T. Bongers
22.5k54460
On two of the three boundaries, the function is just zero. On the last one, you know $x + y = 1$, so you can rewrite your function as a one-variable function
$$f(x, y) = f(x, 1 - x) = x^a e^{-x} (1 - x)^b e^{-(1 - x)} = frac 1 e x^a (1 - x)^b$$
which isn't too hard to maximize.
Then for the interior, try to identify the critical points where $f_x = f_y = 0$.
answered Dec 2 at 21:28
T. Bongers
22.5k54460
answered Dec 2 at 21:28
T. Bongers
22.5k54460
answered Dec 2 at 21:28
T. Bongers
22.5k54460
answered Dec 2 at 21:28
T. Bongers
22.5k54460
22.5k54460
add a comment |
add a comment |
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Lagrange multipliers are only used when you are forced to a constraint in the form of an equality; i.e. "maximise/minimise $f(x)$ given $g(x)=0$"" type questions. You would almost never want to try to introduce $g$ when it's not already there.
– YiFan
Dec 2 at 21:37