Mixing
Finding extrema of a function when one of the variables go away after differentiation
$begingroup$
Sorry for the long title, I wasn't sure how to explain this exactly.
Let's say I have the following function
$f(x,y) = x^2 + 2y$
I want to find the extrema of this function with respect to $x$ and $y$ to try and minimize it. So normally, I would set the partial derivative of the function wrt each variable and set them to zero.
$frac{partial f}{partial x} = 2x$
$2x = 0 Rightarrow x = 0$
So we can say that $x=0$ minimizes this function.
But what can we say about $y$ exactly?
$frac{partial f}{partial y} = 2$
$2= 0$ is nonsensical
Perhaps these types of optimization problems are somehow degenerate?
multivariable-calculus optimization
asked Jan 27 at 6:54
CarpetfizzCarpetfizz
491413
$endgroup$
add a comment |
$begingroup$
Sorry for the long title, I wasn't sure how to explain this exactly.
Let's say I have the following function
$f(x,y) = x^2 + 2y$
I want to find the extrema of this function with respect to $x$ and $y$ to try and minimize it. So normally, I would set the partial derivative of the function wrt each variable and set them to zero.
$frac{partial f}{partial x} = 2x$
$2x = 0 Rightarrow x = 0$
So we can say that $x=0$ minimizes this function.
But what can we say about $y$ exactly?
$frac{partial f}{partial y} = 2$
$2= 0$ is nonsensical
Perhaps these types of optimization problems are somehow degenerate?
multivariable-calculus optimization
asked Jan 27 at 6:54
CarpetfizzCarpetfizz
491413
$endgroup$
2
$begingroup$
This simply means this function is always increasing with regards to the $y$ axis. If you are limited to a closed domain then you can use it's boundaries as your extreme points. If not, then you don't have extreme points.
$endgroup$
– P.Diddy
Jan 27 at 7:16
add a comment |
$begingroup$
Sorry for the long title, I wasn't sure how to explain this exactly.
Let's say I have the following function
$f(x,y) = x^2 + 2y$
I want to find the extrema of this function with respect to $x$ and $y$ to try and minimize it. So normally, I would set the partial derivative of the function wrt each variable and set them to zero.
$frac{partial f}{partial x} = 2x$
$2x = 0 Rightarrow x = 0$
So we can say that $x=0$ minimizes this function.
But what can we say about $y$ exactly?
$frac{partial f}{partial y} = 2$
$2= 0$ is nonsensical
Perhaps these types of optimization problems are somehow degenerate?
multivariable-calculus optimization
asked Jan 27 at 6:54
CarpetfizzCarpetfizz
491413
$endgroup$
Sorry for the long title, I wasn't sure how to explain this exactly.
Let's say I have the following function
$f(x,y) = x^2 + 2y$
I want to find the extrema of this function with respect to $x$ and $y$ to try and minimize it. So normally, I would set the partial derivative of the function wrt each variable and set them to zero.
$frac{partial f}{partial x} = 2x$
$2x = 0 Rightarrow x = 0$
So we can say that $x=0$ minimizes this function.
But what can we say about $y$ exactly?
$frac{partial f}{partial y} = 2$
$2= 0$ is nonsensical
Perhaps these types of optimization problems are somehow degenerate?
multivariable-calculus optimization
multivariable-calculus optimization
asked Jan 27 at 6:54
CarpetfizzCarpetfizz
491413
asked Jan 27 at 6:54
CarpetfizzCarpetfizz
491413
asked Jan 27 at 6:54
CarpetfizzCarpetfizz
491413
asked Jan 27 at 6:54
CarpetfizzCarpetfizz
491413
asked Jan 27 at 6:54
CarpetfizzCarpetfizz
491413
491413
2
$begingroup$
This simply means this function is always increasing with regards to the $y$ axis. If you are limited to a closed domain then you can use it's boundaries as your extreme points. If not, then you don't have extreme points.
$endgroup$
– P.Diddy
Jan 27 at 7:16
add a comment |
2
$begingroup$
This simply means this function is always increasing with regards to the $y$ axis. If you are limited to a closed domain then you can use it's boundaries as your extreme points. If not, then you don't have extreme points.
$endgroup$
– P.Diddy
Jan 27 at 7:16
2
2
$begingroup$
This simply means this function is always increasing with regards to the $y$ axis. If you are limited to a closed domain then you can use it's boundaries as your extreme points. If not, then you don't have extreme points.
$endgroup$
– P.Diddy
Jan 27 at 7:16
$begingroup$
This simply means this function is always increasing with regards to the $y$ axis. If you are limited to a closed domain then you can use it's boundaries as your extreme points. If not, then you don't have extreme points.
$endgroup$
– P.Diddy
Jan 27 at 7:16
add a comment |
1 Answer
1
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oldest
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$begingroup$
$frac{df}{dy}=2$ just means that the derivative is never $0$, so there are no critical points. You can make the function value as small as you want by making $y$ small, so there is no minimum.
answered Jan 27 at 9:06
Erik ParkinsonErik Parkinson
1,17519
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add a comment |
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$begingroup$
$frac{df}{dy}=2$ just means that the derivative is never $0$, so there are no critical points. You can make the function value as small as you want by making $y$ small, so there is no minimum.
answered Jan 27 at 9:06
Erik ParkinsonErik Parkinson
1,17519
$endgroup$
add a comment |
$begingroup$
$frac{df}{dy}=2$ just means that the derivative is never $0$, so there are no critical points. You can make the function value as small as you want by making $y$ small, so there is no minimum.
answered Jan 27 at 9:06
Erik ParkinsonErik Parkinson
1,17519
$endgroup$
add a comment |
$begingroup$
$frac{df}{dy}=2$ just means that the derivative is never $0$, so there are no critical points. You can make the function value as small as you want by making $y$ small, so there is no minimum.
answered Jan 27 at 9:06
Erik ParkinsonErik Parkinson
1,17519
$endgroup$
$frac{df}{dy}=2$ just means that the derivative is never $0$, so there are no critical points. You can make the function value as small as you want by making $y$ small, so there is no minimum.
answered Jan 27 at 9:06
Erik ParkinsonErik Parkinson
1,17519
answered Jan 27 at 9:06
Erik ParkinsonErik Parkinson
1,17519
answered Jan 27 at 9:06
Erik ParkinsonErik Parkinson
1,17519
answered Jan 27 at 9:06
Erik ParkinsonErik Parkinson
1,17519
1,17519
add a comment |
add a comment |
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$begingroup$
This simply means this function is always increasing with regards to the $y$ axis. If you are limited to a closed domain then you can use it's boundaries as your extreme points. If not, then you don't have extreme points.
$endgroup$
– P.Diddy
Jan 27 at 7:16