Finding the Maximum and Minimum Values of a Function in a Domain
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Find the maximum and minimum values of the function $f(x, y) = 2x^2+3y^2-4x-5$ on the domain $x^2+y^2le 225$.
After finding the first partial derivatives, I found that $(1, 0)$ was a critical point and I found that it was a local minimum from the second derivative test. So the minimum value of $f(x, y)$ would be $-7$ at the point $(1, 0)$.
However, what I am confused about is how to find the maximum value and point. Since this function does not have a local maximum point, I thought that the answer would simply be on the boundaries of the inequality. However, it seems that neither the point $(15, 0)$ or $(0, 15)$ give the correct answer.
If anyone knows how I should approach this problem and can provide feedback, I would be very grateful!
calculus multivariable-calculus optimization partial-derivative
asked Nov 24 '18 at 6:07
sktsasussktsasus
1,027519
$endgroup$
add a comment |
$begingroup$
Find the maximum and minimum values of the function $f(x, y) = 2x^2+3y^2-4x-5$ on the domain $x^2+y^2le 225$.
After finding the first partial derivatives, I found that $(1, 0)$ was a critical point and I found that it was a local minimum from the second derivative test. So the minimum value of $f(x, y)$ would be $-7$ at the point $(1, 0)$.
However, what I am confused about is how to find the maximum value and point. Since this function does not have a local maximum point, I thought that the answer would simply be on the boundaries of the inequality. However, it seems that neither the point $(15, 0)$ or $(0, 15)$ give the correct answer.
If anyone knows how I should approach this problem and can provide feedback, I would be very grateful!
calculus multivariable-calculus optimization partial-derivative
asked Nov 24 '18 at 6:07
sktsasussktsasus
1,027519
$endgroup$
1
$begingroup$
The boundary of the domain is a full circle defined by $x^2+y^2leq 225$, not just the two points $(15,0)$ and $(0,15)$.
$endgroup$
– 高田航
Nov 24 '18 at 6:15
$begingroup$
Yes I see. So in that case, could you please tell me how I should approach this problem instead? Thank you!
$endgroup$
– sktsasus
Nov 24 '18 at 6:31
$begingroup$
have you heard about Langrange multipliers?
$endgroup$
– Thomas
Nov 24 '18 at 6:37
add a comment |
$begingroup$
Find the maximum and minimum values of the function $f(x, y) = 2x^2+3y^2-4x-5$ on the domain $x^2+y^2le 225$.
After finding the first partial derivatives, I found that $(1, 0)$ was a critical point and I found that it was a local minimum from the second derivative test. So the minimum value of $f(x, y)$ would be $-7$ at the point $(1, 0)$.
However, what I am confused about is how to find the maximum value and point. Since this function does not have a local maximum point, I thought that the answer would simply be on the boundaries of the inequality. However, it seems that neither the point $(15, 0)$ or $(0, 15)$ give the correct answer.
If anyone knows how I should approach this problem and can provide feedback, I would be very grateful!
calculus multivariable-calculus optimization partial-derivative
asked Nov 24 '18 at 6:07
sktsasussktsasus
1,027519
$endgroup$
Find the maximum and minimum values of the function $f(x, y) = 2x^2+3y^2-4x-5$ on the domain $x^2+y^2le 225$.
After finding the first partial derivatives, I found that $(1, 0)$ was a critical point and I found that it was a local minimum from the second derivative test. So the minimum value of $f(x, y)$ would be $-7$ at the point $(1, 0)$.
However, what I am confused about is how to find the maximum value and point. Since this function does not have a local maximum point, I thought that the answer would simply be on the boundaries of the inequality. However, it seems that neither the point $(15, 0)$ or $(0, 15)$ give the correct answer.
If anyone knows how I should approach this problem and can provide feedback, I would be very grateful!
calculus multivariable-calculus optimization partial-derivative
calculus multivariable-calculus optimization partial-derivative
asked Nov 24 '18 at 6:07
sktsasussktsasus
1,027519
asked Nov 24 '18 at 6:07
sktsasussktsasus
1,027519
asked Nov 24 '18 at 6:07
sktsasussktsasus
1,027519
asked Nov 24 '18 at 6:07
sktsasussktsasus
1,027519
asked Nov 24 '18 at 6:07
sktsasussktsasus
1,027519
1,027519
1
$begingroup$
The boundary of the domain is a full circle defined by $x^2+y^2leq 225$, not just the two points $(15,0)$ and $(0,15)$.
$endgroup$
– 高田航
Nov 24 '18 at 6:15
$begingroup$
Yes I see. So in that case, could you please tell me how I should approach this problem instead? Thank you!
$endgroup$
– sktsasus
Nov 24 '18 at 6:31
$begingroup$
have you heard about Langrange multipliers?
$endgroup$
– Thomas
Nov 24 '18 at 6:37
add a comment |
1
$begingroup$
The boundary of the domain is a full circle defined by $x^2+y^2leq 225$, not just the two points $(15,0)$ and $(0,15)$.
$endgroup$
– 高田航
Nov 24 '18 at 6:15
$begingroup$
Yes I see. So in that case, could you please tell me how I should approach this problem instead? Thank you!
$endgroup$
– sktsasus
Nov 24 '18 at 6:31
$begingroup$
have you heard about Langrange multipliers?
$endgroup$
– Thomas
Nov 24 '18 at 6:37
1
1
$begingroup$
The boundary of the domain is a full circle defined by $x^2+y^2leq 225$, not just the two points $(15,0)$ and $(0,15)$.
$endgroup$
– 高田航
Nov 24 '18 at 6:15
$begingroup$
The boundary of the domain is a full circle defined by $x^2+y^2leq 225$, not just the two points $(15,0)$ and $(0,15)$.
$endgroup$
– 高田航
Nov 24 '18 at 6:15
$begingroup$
Yes I see. So in that case, could you please tell me how I should approach this problem instead? Thank you!
$endgroup$
– sktsasus
Nov 24 '18 at 6:31
$begingroup$
Yes I see. So in that case, could you please tell me how I should approach this problem instead? Thank you!
$endgroup$
– sktsasus
Nov 24 '18 at 6:31
$begingroup$
have you heard about Langrange multipliers?
$endgroup$
– Thomas
Nov 24 '18 at 6:37
$begingroup$
have you heard about Langrange multipliers?
$endgroup$
– Thomas
Nov 24 '18 at 6:37
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
According to the method of Lagrange parameters, the gradient of $f$ will be normal to the boundary in an extremal value. The normal to the circle in a given point $(x,y)$ is just the position vector $(x,y)$, so you have to look for points where $nabla f$ is a mulitple of that vector.
Now
$$nabla f =left ( array{4x-4 \ 6y} right) $$
which means we are looking for $x, y$ such that $ 4x-4 = lambda x, 6y= lambda y$ with the side condition $x^2+y^2=225$
This is true iff ($lambda = 6$ or $y=0$).
If $lambda = 6$ you easily see that then $x=-2$ and $y$ is determined by the side condition.
If $y=0$ $x$ is determined by the side condition (and $lambda$ by the equation for $x$ and $lambda$).
Plugging in $x$ and $y$ into the function will show whether you have a maximum or minimum.
answered Nov 24 '18 at 6:49
ThomasThomas
16.9k21731
$endgroup$
$begingroup$
Thank you for the answer! I am familiar with Lagrange Multipliers but I just wasn't sure how they would work here. But now I do see that I needed to check for the global minimum as well as the critical point I found may not have been it. Thanks for the clarification!
$endgroup$
– sktsasus
Nov 24 '18 at 16:08
add a comment |
$begingroup$
As the guy in the comment says, the boundary of the domain is the circumference of the circle $x^2 + y^2 = 225$, not just the two points. Since the function has a minima at $(1,0)$ and no other critical points in the region, it means that the function is increasing in every direction from that point. So the maxima of the function will be at one of the boundary points. Now if the point is on the boundary, it will lie on the circle's circumference and satisfy the equation $x^2 + y^2 = 225$. Putting this in the function equation, we get
$$f(x,y) = 2x^2 + 3y^2 -4x -5$$
$$2x^2 + 3(225 - x^2) -4x -5$$
$$g(x)=-x^2 -4x + 670$$
Now you can diffrentiate to find out where this will have its maxima
$$g'(x) = -2x - 4 = 0$$
$$x=-2$$
As you can quickly confirm from the original equation or the graph of a quadratic equation, this is a maxima. Using this to get $y$
$$y^2 = 225 - x^2 = 225 -4 = 221$$
$$y = sqrt{221}$$
Putting $x=-2$ and $y=sqrt{221}$ back in the original function
$$f(x,y) = 674$$
answered Nov 24 '18 at 6:36
Sauhard SharmaSauhard Sharma
953318
$endgroup$
1
$begingroup$
Note the function must either be constant or must have at least one minimum and one maximum along the boundary, since that is compact and the function is continuous. You are loosing at least one solution by your approach.
$endgroup$
– Thomas
Nov 24 '18 at 6:55
2
$begingroup$
Actually, $f(x,y) = f(x, -y)$, so with your values of $x, y$ you found two minima or maxima with the same value. This means you actually missed at least two other extrem values.
$endgroup$
– Thomas
Nov 24 '18 at 7:32
1
$begingroup$
Hint: note that $-25 le xle 25$, so also the one dimensional problem to which you reduced the two dimensional one is an extreme value problem with a boundary condition.
$endgroup$
– Thomas
Nov 24 '18 at 7:34
2
$begingroup$
The values you found for $y$ are the locations of the two maxima. In $y=0, x=pm 25$ you have two more extreme values (with respect to the boundary), which are the minima of $f$ along the boundary (but not global minima, the global minimum is attained in the interior of the region, which was discoverd by the OP already).
$endgroup$
– Thomas
Nov 24 '18 at 10:04
1
$begingroup$
@Thomas Every occurrence of "25" should be "15" in your comments. This doesn't affect anything else, but may have misled others.
$endgroup$
– Teepeemm
Nov 24 '18 at 16:07
|
show 5 more comments
$begingroup$
If you don't know, or want to use, Lagrange Multipliers as in Thomas's answer, and you want to find the local extrema on the boundary to make sure you don't miss anything, then you can either handle the endpoints with $x=pm15$ that Sauhard Sharma neglected to address in their answer, or you can give the circle a more natural parametrization.
Circles can be parametrized with cosines and sines, so the circle $x^2+y^2=225$ is parametrized by $x=15cos t,, y=15sin t$. If you substitute this in to the objective function $2x^2+3y^2-4x-5$, you'll see how the values of the function above the circle depend on the angle $t$, and you can then use single-variable calculus to find the local extrema. Since cosine and sine are continuous on the whole real line, you don't have to worry about missing end-points with this method.
answered Nov 24 '18 at 14:34
Mark S.Mark S.
12.4k22772
$endgroup$
add a comment |
$begingroup$
With free CAS Maxima and my programm "nopt":
package "nopt"
examples
answered Nov 24 '18 at 6:51
Aleksas DomarkasAleksas Domarkas
1,62317
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@ Mark S Link to package "nopt" I include.
$endgroup$
– Aleksas Domarkas
Nov 26 '18 at 11:29
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
According to the method of Lagrange parameters, the gradient of $f$ will be normal to the boundary in an extremal value. The normal to the circle in a given point $(x,y)$ is just the position vector $(x,y)$, so you have to look for points where $nabla f$ is a mulitple of that vector.
Now
$$nabla f =left ( array{4x-4 \ 6y} right) $$
which means we are looking for $x, y$ such that $ 4x-4 = lambda x, 6y= lambda y$ with the side condition $x^2+y^2=225$
This is true iff ($lambda = 6$ or $y=0$).
If $lambda = 6$ you easily see that then $x=-2$ and $y$ is determined by the side condition.
If $y=0$ $x$ is determined by the side condition (and $lambda$ by the equation for $x$ and $lambda$).
Plugging in $x$ and $y$ into the function will show whether you have a maximum or minimum.
answered Nov 24 '18 at 6:49
ThomasThomas
16.9k21731
$endgroup$
$begingroup$
Thank you for the answer! I am familiar with Lagrange Multipliers but I just wasn't sure how they would work here. But now I do see that I needed to check for the global minimum as well as the critical point I found may not have been it. Thanks for the clarification!
$endgroup$
– sktsasus
Nov 24 '18 at 16:08
add a comment |
$begingroup$
According to the method of Lagrange parameters, the gradient of $f$ will be normal to the boundary in an extremal value. The normal to the circle in a given point $(x,y)$ is just the position vector $(x,y)$, so you have to look for points where $nabla f$ is a mulitple of that vector.
Now
$$nabla f =left ( array{4x-4 \ 6y} right) $$
which means we are looking for $x, y$ such that $ 4x-4 = lambda x, 6y= lambda y$ with the side condition $x^2+y^2=225$
This is true iff ($lambda = 6$ or $y=0$).
If $lambda = 6$ you easily see that then $x=-2$ and $y$ is determined by the side condition.
If $y=0$ $x$ is determined by the side condition (and $lambda$ by the equation for $x$ and $lambda$).
Plugging in $x$ and $y$ into the function will show whether you have a maximum or minimum.
answered Nov 24 '18 at 6:49
ThomasThomas
16.9k21731
$endgroup$
$begingroup$
Thank you for the answer! I am familiar with Lagrange Multipliers but I just wasn't sure how they would work here. But now I do see that I needed to check for the global minimum as well as the critical point I found may not have been it. Thanks for the clarification!
$endgroup$
– sktsasus
Nov 24 '18 at 16:08
add a comment |
$begingroup$
According to the method of Lagrange parameters, the gradient of $f$ will be normal to the boundary in an extremal value. The normal to the circle in a given point $(x,y)$ is just the position vector $(x,y)$, so you have to look for points where $nabla f$ is a mulitple of that vector.
Now
$$nabla f =left ( array{4x-4 \ 6y} right) $$
which means we are looking for $x, y$ such that $ 4x-4 = lambda x, 6y= lambda y$ with the side condition $x^2+y^2=225$
This is true iff ($lambda = 6$ or $y=0$).
If $lambda = 6$ you easily see that then $x=-2$ and $y$ is determined by the side condition.
If $y=0$ $x$ is determined by the side condition (and $lambda$ by the equation for $x$ and $lambda$).
Plugging in $x$ and $y$ into the function will show whether you have a maximum or minimum.
answered Nov 24 '18 at 6:49
ThomasThomas
16.9k21731
$endgroup$
According to the method of Lagrange parameters, the gradient of $f$ will be normal to the boundary in an extremal value. The normal to the circle in a given point $(x,y)$ is just the position vector $(x,y)$, so you have to look for points where $nabla f$ is a mulitple of that vector.
Now
$$nabla f =left ( array{4x-4 \ 6y} right) $$
which means we are looking for $x, y$ such that $ 4x-4 = lambda x, 6y= lambda y$ with the side condition $x^2+y^2=225$
This is true iff ($lambda = 6$ or $y=0$).
If $lambda = 6$ you easily see that then $x=-2$ and $y$ is determined by the side condition.
If $y=0$ $x$ is determined by the side condition (and $lambda$ by the equation for $x$ and $lambda$).
Plugging in $x$ and $y$ into the function will show whether you have a maximum or minimum.
answered Nov 24 '18 at 6:49
ThomasThomas
16.9k21731
edited Nov 24 '18 at 7:34
answered Nov 24 '18 at 6:49
ThomasThomas
16.9k21731
answered Nov 24 '18 at 6:49
ThomasThomas
16.9k21731
answered Nov 24 '18 at 6:49
ThomasThomas
16.9k21731
16.9k21731
$begingroup$
Thank you for the answer! I am familiar with Lagrange Multipliers but I just wasn't sure how they would work here. But now I do see that I needed to check for the global minimum as well as the critical point I found may not have been it. Thanks for the clarification!
$endgroup$
– sktsasus
Nov 24 '18 at 16:08
add a comment |
$begingroup$
Thank you for the answer! I am familiar with Lagrange Multipliers but I just wasn't sure how they would work here. But now I do see that I needed to check for the global minimum as well as the critical point I found may not have been it. Thanks for the clarification!
$endgroup$
– sktsasus
Nov 24 '18 at 16:08
$begingroup$
Thank you for the answer! I am familiar with Lagrange Multipliers but I just wasn't sure how they would work here. But now I do see that I needed to check for the global minimum as well as the critical point I found may not have been it. Thanks for the clarification!
$endgroup$
– sktsasus
Nov 24 '18 at 16:08
$begingroup$
Thank you for the answer! I am familiar with Lagrange Multipliers but I just wasn't sure how they would work here. But now I do see that I needed to check for the global minimum as well as the critical point I found may not have been it. Thanks for the clarification!
$endgroup$
– sktsasus
Nov 24 '18 at 16:08
add a comment |
$begingroup$
As the guy in the comment says, the boundary of the domain is the circumference of the circle $x^2 + y^2 = 225$, not just the two points. Since the function has a minima at $(1,0)$ and no other critical points in the region, it means that the function is increasing in every direction from that point. So the maxima of the function will be at one of the boundary points. Now if the point is on the boundary, it will lie on the circle's circumference and satisfy the equation $x^2 + y^2 = 225$. Putting this in the function equation, we get
$$f(x,y) = 2x^2 + 3y^2 -4x -5$$
$$2x^2 + 3(225 - x^2) -4x -5$$
$$g(x)=-x^2 -4x + 670$$
Now you can diffrentiate to find out where this will have its maxima
$$g'(x) = -2x - 4 = 0$$
$$x=-2$$
As you can quickly confirm from the original equation or the graph of a quadratic equation, this is a maxima. Using this to get $y$
$$y^2 = 225 - x^2 = 225 -4 = 221$$
$$y = sqrt{221}$$
Putting $x=-2$ and $y=sqrt{221}$ back in the original function
$$f(x,y) = 674$$
answered Nov 24 '18 at 6:36
Sauhard SharmaSauhard Sharma
953318
$endgroup$
1
$begingroup$
Note the function must either be constant or must have at least one minimum and one maximum along the boundary, since that is compact and the function is continuous. You are loosing at least one solution by your approach.
$endgroup$
– Thomas
Nov 24 '18 at 6:55
2
$begingroup$
Actually, $f(x,y) = f(x, -y)$, so with your values of $x, y$ you found two minima or maxima with the same value. This means you actually missed at least two other extrem values.
$endgroup$
– Thomas
Nov 24 '18 at 7:32
1
$begingroup$
Hint: note that $-25 le xle 25$, so also the one dimensional problem to which you reduced the two dimensional one is an extreme value problem with a boundary condition.
$endgroup$
– Thomas
Nov 24 '18 at 7:34
2
$begingroup$
The values you found for $y$ are the locations of the two maxima. In $y=0, x=pm 25$ you have two more extreme values (with respect to the boundary), which are the minima of $f$ along the boundary (but not global minima, the global minimum is attained in the interior of the region, which was discoverd by the OP already).
$endgroup$
– Thomas
Nov 24 '18 at 10:04
1
$begingroup$
@Thomas Every occurrence of "25" should be "15" in your comments. This doesn't affect anything else, but may have misled others.
$endgroup$
– Teepeemm
Nov 24 '18 at 16:07
|
show 5 more comments
$begingroup$
As the guy in the comment says, the boundary of the domain is the circumference of the circle $x^2 + y^2 = 225$, not just the two points. Since the function has a minima at $(1,0)$ and no other critical points in the region, it means that the function is increasing in every direction from that point. So the maxima of the function will be at one of the boundary points. Now if the point is on the boundary, it will lie on the circle's circumference and satisfy the equation $x^2 + y^2 = 225$. Putting this in the function equation, we get
$$f(x,y) = 2x^2 + 3y^2 -4x -5$$
$$2x^2 + 3(225 - x^2) -4x -5$$
$$g(x)=-x^2 -4x + 670$$
Now you can diffrentiate to find out where this will have its maxima
$$g'(x) = -2x - 4 = 0$$
$$x=-2$$
As you can quickly confirm from the original equation or the graph of a quadratic equation, this is a maxima. Using this to get $y$
$$y^2 = 225 - x^2 = 225 -4 = 221$$
$$y = sqrt{221}$$
Putting $x=-2$ and $y=sqrt{221}$ back in the original function
$$f(x,y) = 674$$
answered Nov 24 '18 at 6:36
Sauhard SharmaSauhard Sharma
953318
$endgroup$
1
$begingroup$
Note the function must either be constant or must have at least one minimum and one maximum along the boundary, since that is compact and the function is continuous. You are loosing at least one solution by your approach.
$endgroup$
– Thomas
Nov 24 '18 at 6:55
2
$begingroup$
Actually, $f(x,y) = f(x, -y)$, so with your values of $x, y$ you found two minima or maxima with the same value. This means you actually missed at least two other extrem values.
$endgroup$
– Thomas
Nov 24 '18 at 7:32
1
$begingroup$
Hint: note that $-25 le xle 25$, so also the one dimensional problem to which you reduced the two dimensional one is an extreme value problem with a boundary condition.
$endgroup$
– Thomas
Nov 24 '18 at 7:34
2
$begingroup$
The values you found for $y$ are the locations of the two maxima. In $y=0, x=pm 25$ you have two more extreme values (with respect to the boundary), which are the minima of $f$ along the boundary (but not global minima, the global minimum is attained in the interior of the region, which was discoverd by the OP already).
$endgroup$
– Thomas
Nov 24 '18 at 10:04
1
$begingroup$
@Thomas Every occurrence of "25" should be "15" in your comments. This doesn't affect anything else, but may have misled others.
$endgroup$
– Teepeemm
Nov 24 '18 at 16:07
|
show 5 more comments
$begingroup$
As the guy in the comment says, the boundary of the domain is the circumference of the circle $x^2 + y^2 = 225$, not just the two points. Since the function has a minima at $(1,0)$ and no other critical points in the region, it means that the function is increasing in every direction from that point. So the maxima of the function will be at one of the boundary points. Now if the point is on the boundary, it will lie on the circle's circumference and satisfy the equation $x^2 + y^2 = 225$. Putting this in the function equation, we get
$$f(x,y) = 2x^2 + 3y^2 -4x -5$$
$$2x^2 + 3(225 - x^2) -4x -5$$
$$g(x)=-x^2 -4x + 670$$
Now you can diffrentiate to find out where this will have its maxima
$$g'(x) = -2x - 4 = 0$$
$$x=-2$$
As you can quickly confirm from the original equation or the graph of a quadratic equation, this is a maxima. Using this to get $y$
$$y^2 = 225 - x^2 = 225 -4 = 221$$
$$y = sqrt{221}$$
Putting $x=-2$ and $y=sqrt{221}$ back in the original function
$$f(x,y) = 674$$
answered Nov 24 '18 at 6:36
Sauhard SharmaSauhard Sharma
953318
$endgroup$
As the guy in the comment says, the boundary of the domain is the circumference of the circle $x^2 + y^2 = 225$, not just the two points. Since the function has a minima at $(1,0)$ and no other critical points in the region, it means that the function is increasing in every direction from that point. So the maxima of the function will be at one of the boundary points. Now if the point is on the boundary, it will lie on the circle's circumference and satisfy the equation $x^2 + y^2 = 225$. Putting this in the function equation, we get
$$f(x,y) = 2x^2 + 3y^2 -4x -5$$
$$2x^2 + 3(225 - x^2) -4x -5$$
$$g(x)=-x^2 -4x + 670$$
Now you can diffrentiate to find out where this will have its maxima
$$g'(x) = -2x - 4 = 0$$
$$x=-2$$
As you can quickly confirm from the original equation or the graph of a quadratic equation, this is a maxima. Using this to get $y$
$$y^2 = 225 - x^2 = 225 -4 = 221$$
$$y = sqrt{221}$$
Putting $x=-2$ and $y=sqrt{221}$ back in the original function
$$f(x,y) = 674$$
answered Nov 24 '18 at 6:36
Sauhard SharmaSauhard Sharma
953318
answered Nov 24 '18 at 6:36
Sauhard SharmaSauhard Sharma
953318
answered Nov 24 '18 at 6:36
Sauhard SharmaSauhard Sharma
953318
answered Nov 24 '18 at 6:36
Sauhard SharmaSauhard Sharma
953318
953318
1
$begingroup$
Note the function must either be constant or must have at least one minimum and one maximum along the boundary, since that is compact and the function is continuous. You are loosing at least one solution by your approach.
$endgroup$
– Thomas
Nov 24 '18 at 6:55
2
$begingroup$
Actually, $f(x,y) = f(x, -y)$, so with your values of $x, y$ you found two minima or maxima with the same value. This means you actually missed at least two other extrem values.
$endgroup$
– Thomas
Nov 24 '18 at 7:32
1
$begingroup$
Hint: note that $-25 le xle 25$, so also the one dimensional problem to which you reduced the two dimensional one is an extreme value problem with a boundary condition.
$endgroup$
– Thomas
Nov 24 '18 at 7:34
2
$begingroup$
The values you found for $y$ are the locations of the two maxima. In $y=0, x=pm 25$ you have two more extreme values (with respect to the boundary), which are the minima of $f$ along the boundary (but not global minima, the global minimum is attained in the interior of the region, which was discoverd by the OP already).
$endgroup$
– Thomas
Nov 24 '18 at 10:04
1
$begingroup$
@Thomas Every occurrence of "25" should be "15" in your comments. This doesn't affect anything else, but may have misled others.
$endgroup$
– Teepeemm
Nov 24 '18 at 16:07
|
show 5 more comments
1
$begingroup$
Note the function must either be constant or must have at least one minimum and one maximum along the boundary, since that is compact and the function is continuous. You are loosing at least one solution by your approach.
$endgroup$
– Thomas
Nov 24 '18 at 6:55
2
$begingroup$
Actually, $f(x,y) = f(x, -y)$, so with your values of $x, y$ you found two minima or maxima with the same value. This means you actually missed at least two other extrem values.
$endgroup$
– Thomas
Nov 24 '18 at 7:32
1
$begingroup$
Hint: note that $-25 le xle 25$, so also the one dimensional problem to which you reduced the two dimensional one is an extreme value problem with a boundary condition.
$endgroup$
– Thomas
Nov 24 '18 at 7:34
2
$begingroup$
The values you found for $y$ are the locations of the two maxima. In $y=0, x=pm 25$ you have two more extreme values (with respect to the boundary), which are the minima of $f$ along the boundary (but not global minima, the global minimum is attained in the interior of the region, which was discoverd by the OP already).
$endgroup$
– Thomas
Nov 24 '18 at 10:04
1
$begingroup$
@Thomas Every occurrence of "25" should be "15" in your comments. This doesn't affect anything else, but may have misled others.
$endgroup$
– Teepeemm
Nov 24 '18 at 16:07
1
1
$begingroup$
Note the function must either be constant or must have at least one minimum and one maximum along the boundary, since that is compact and the function is continuous. You are loosing at least one solution by your approach.
$endgroup$
– Thomas
Nov 24 '18 at 6:55
$begingroup$
Note the function must either be constant or must have at least one minimum and one maximum along the boundary, since that is compact and the function is continuous. You are loosing at least one solution by your approach.
$endgroup$
– Thomas
Nov 24 '18 at 6:55
2
2
$begingroup$
Actually, $f(x,y) = f(x, -y)$, so with your values of $x, y$ you found two minima or maxima with the same value. This means you actually missed at least two other extrem values.
$endgroup$
– Thomas
Nov 24 '18 at 7:32
$begingroup$
Actually, $f(x,y) = f(x, -y)$, so with your values of $x, y$ you found two minima or maxima with the same value. This means you actually missed at least two other extrem values.
$endgroup$
– Thomas
Nov 24 '18 at 7:32
1
1
$begingroup$
Hint: note that $-25 le xle 25$, so also the one dimensional problem to which you reduced the two dimensional one is an extreme value problem with a boundary condition.
$endgroup$
– Thomas
Nov 24 '18 at 7:34
$begingroup$
Hint: note that $-25 le xle 25$, so also the one dimensional problem to which you reduced the two dimensional one is an extreme value problem with a boundary condition.
$endgroup$
– Thomas
Nov 24 '18 at 7:34
2
2
$begingroup$
The values you found for $y$ are the locations of the two maxima. In $y=0, x=pm 25$ you have two more extreme values (with respect to the boundary), which are the minima of $f$ along the boundary (but not global minima, the global minimum is attained in the interior of the region, which was discoverd by the OP already).
$endgroup$
– Thomas
Nov 24 '18 at 10:04
$begingroup$
The values you found for $y$ are the locations of the two maxima. In $y=0, x=pm 25$ you have two more extreme values (with respect to the boundary), which are the minima of $f$ along the boundary (but not global minima, the global minimum is attained in the interior of the region, which was discoverd by the OP already).
$endgroup$
– Thomas
Nov 24 '18 at 10:04
1
1
$begingroup$
@Thomas Every occurrence of "25" should be "15" in your comments. This doesn't affect anything else, but may have misled others.
$endgroup$
– Teepeemm
Nov 24 '18 at 16:07
$begingroup$
@Thomas Every occurrence of "25" should be "15" in your comments. This doesn't affect anything else, but may have misled others.
$endgroup$
– Teepeemm
Nov 24 '18 at 16:07
|
show 5 more comments
$begingroup$
If you don't know, or want to use, Lagrange Multipliers as in Thomas's answer, and you want to find the local extrema on the boundary to make sure you don't miss anything, then you can either handle the endpoints with $x=pm15$ that Sauhard Sharma neglected to address in their answer, or you can give the circle a more natural parametrization.
Circles can be parametrized with cosines and sines, so the circle $x^2+y^2=225$ is parametrized by $x=15cos t,, y=15sin t$. If you substitute this in to the objective function $2x^2+3y^2-4x-5$, you'll see how the values of the function above the circle depend on the angle $t$, and you can then use single-variable calculus to find the local extrema. Since cosine and sine are continuous on the whole real line, you don't have to worry about missing end-points with this method.
answered Nov 24 '18 at 14:34
Mark S.Mark S.
12.4k22772
$endgroup$
add a comment |
$begingroup$
If you don't know, or want to use, Lagrange Multipliers as in Thomas's answer, and you want to find the local extrema on the boundary to make sure you don't miss anything, then you can either handle the endpoints with $x=pm15$ that Sauhard Sharma neglected to address in their answer, or you can give the circle a more natural parametrization.
Circles can be parametrized with cosines and sines, so the circle $x^2+y^2=225$ is parametrized by $x=15cos t,, y=15sin t$. If you substitute this in to the objective function $2x^2+3y^2-4x-5$, you'll see how the values of the function above the circle depend on the angle $t$, and you can then use single-variable calculus to find the local extrema. Since cosine and sine are continuous on the whole real line, you don't have to worry about missing end-points with this method.
answered Nov 24 '18 at 14:34
Mark S.Mark S.
12.4k22772
$endgroup$
add a comment |
$begingroup$
If you don't know, or want to use, Lagrange Multipliers as in Thomas's answer, and you want to find the local extrema on the boundary to make sure you don't miss anything, then you can either handle the endpoints with $x=pm15$ that Sauhard Sharma neglected to address in their answer, or you can give the circle a more natural parametrization.
Circles can be parametrized with cosines and sines, so the circle $x^2+y^2=225$ is parametrized by $x=15cos t,, y=15sin t$. If you substitute this in to the objective function $2x^2+3y^2-4x-5$, you'll see how the values of the function above the circle depend on the angle $t$, and you can then use single-variable calculus to find the local extrema. Since cosine and sine are continuous on the whole real line, you don't have to worry about missing end-points with this method.
answered Nov 24 '18 at 14:34
Mark S.Mark S.
12.4k22772
$endgroup$
If you don't know, or want to use, Lagrange Multipliers as in Thomas's answer, and you want to find the local extrema on the boundary to make sure you don't miss anything, then you can either handle the endpoints with $x=pm15$ that Sauhard Sharma neglected to address in their answer, or you can give the circle a more natural parametrization.
Circles can be parametrized with cosines and sines, so the circle $x^2+y^2=225$ is parametrized by $x=15cos t,, y=15sin t$. If you substitute this in to the objective function $2x^2+3y^2-4x-5$, you'll see how the values of the function above the circle depend on the angle $t$, and you can then use single-variable calculus to find the local extrema. Since cosine and sine are continuous on the whole real line, you don't have to worry about missing end-points with this method.
answered Nov 24 '18 at 14:34
Mark S.Mark S.
12.4k22772
answered Nov 24 '18 at 14:34
Mark S.Mark S.
12.4k22772
answered Nov 24 '18 at 14:34
Mark S.Mark S.
12.4k22772
answered Nov 24 '18 at 14:34
Mark S.Mark S.
12.4k22772
12.4k22772
add a comment |
add a comment |
$begingroup$
With free CAS Maxima and my programm "nopt":
package "nopt"
examples
answered Nov 24 '18 at 6:51
Aleksas DomarkasAleksas Domarkas
1,62317
$endgroup$
$begingroup$
@ Mark S Link to package "nopt" I include.
$endgroup$
– Aleksas Domarkas
Nov 26 '18 at 11:29
add a comment |
$begingroup$
With free CAS Maxima and my programm "nopt":
package "nopt"
examples
answered Nov 24 '18 at 6:51
Aleksas DomarkasAleksas Domarkas
1,62317
$endgroup$
$begingroup$
@ Mark S Link to package "nopt" I include.
$endgroup$
– Aleksas Domarkas
Nov 26 '18 at 11:29
add a comment |
$begingroup$
With free CAS Maxima and my programm "nopt":
package "nopt"
examples
answered Nov 24 '18 at 6:51
Aleksas DomarkasAleksas Domarkas
1,62317
$endgroup$
With free CAS Maxima and my programm "nopt":
package "nopt"
examples
answered Nov 24 '18 at 6:51
Aleksas DomarkasAleksas Domarkas
1,62317
edited Nov 26 '18 at 11:22
answered Nov 24 '18 at 6:51
Aleksas DomarkasAleksas Domarkas
1,62317
answered Nov 24 '18 at 6:51
Aleksas DomarkasAleksas Domarkas
1,62317
answered Nov 24 '18 at 6:51
Aleksas DomarkasAleksas Domarkas
1,62317
1,62317
$begingroup$
@ Mark S Link to package "nopt" I include.
$endgroup$
– Aleksas Domarkas
Nov 26 '18 at 11:29
add a comment |
$begingroup$
@ Mark S Link to package "nopt" I include.
$endgroup$
– Aleksas Domarkas
Nov 26 '18 at 11:29
$begingroup$
@ Mark S Link to package "nopt" I include.
$endgroup$
– Aleksas Domarkas
Nov 26 '18 at 11:29
$begingroup$
@ Mark S Link to package "nopt" I include.
$endgroup$
– Aleksas Domarkas
Nov 26 '18 at 11:29
add a comment |
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$begingroup$
The boundary of the domain is a full circle defined by $x^2+y^2leq 225$, not just the two points $(15,0)$ and $(0,15)$.
$endgroup$
– 高田航
Nov 24 '18 at 6:15
$begingroup$
Yes I see. So in that case, could you please tell me how I should approach this problem instead? Thank you!
$endgroup$
– sktsasus
Nov 24 '18 at 6:31
$begingroup$
have you heard about Langrange multipliers?
$endgroup$
– Thomas
Nov 24 '18 at 6:37