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Karush-Kuhn-Tucker equation [closed]
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Consider the following maximazation problem
Maximize $x^2+y-1$
Subject to $x^2+y^2<1$
Required...
1) Write Karush-Kuhn-Tucker equation that are needed to identify the extreme of $F$ on $A$
2) Determine the physical point that satisfy the Karush-Kuhn-Tucker's equation
economics
edited Aug 1 at 20:21
bjcolby15
7961616
asked Aug 1 at 17:54
RudY the Heck
42
closed as off-topic by Isaac Browne, Ethan Bolker, mlc, Math Lover, callculus Aug 1 at 18:06
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Isaac Browne, Ethan Bolker, mlc, Math Lover, callculus
add a comment |Â
up vote
0
down vote
favorite
Consider the following maximazation problem
Maximize $x^2+y-1$
Subject to $x^2+y^2<1$
Required...
1) Write Karush-Kuhn-Tucker equation that are needed to identify the extreme of $F$ on $A$
2) Determine the physical point that satisfy the Karush-Kuhn-Tucker's equation
economics
edited Aug 1 at 20:21
bjcolby15
7961616
asked Aug 1 at 17:54
RudY the Heck
42
closed as off-topic by Isaac Browne, Ethan Bolker, mlc, Math Lover, callculus Aug 1 at 18:06
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Isaac Browne, Ethan Bolker, mlc, Math Lover, callculus
3
Kuhn tuckey, kuhn taken, kuhn takers...
– Kenny Lau
Aug 1 at 17:55
Kuhn turkey, tucker, ticker
– amWhy
Aug 1 at 18:04
Ok it's to say no idea rather than spittng such ......
– RudY the Heck
Aug 1 at 18:23
Follow this. The function $f(x,y)=x^2+y-1$. The function $g(x,y)=x^2+y^2-1$, which defines the restriction $g(x,y)leq0$ equivalent to $x^2+y^2leq1$. Notice that the restriction must have $leq$ and not $<$. Otherwise there is no maximum. Then $nabla f(x,y)=(2x,1)$, $nabla g(x,y)=(2x,2y)$. At an maximum $(x_0,y_0)$ we must have $(2x_0,1)=mu (2x_0,2y_0)$, with $x_0^2+y_0^2leq 1$, $mu leq 0$ and $mu(x_0^2+y_0^2-1)=0$.
– spiralstotheleft
Aug 2 at 2:03
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Consider the following maximazation problem
Maximize $x^2+y-1$
Subject to $x^2+y^2<1$
Required...
1) Write Karush-Kuhn-Tucker equation that are needed to identify the extreme of $F$ on $A$
2) Determine the physical point that satisfy the Karush-Kuhn-Tucker's equation
economics
edited Aug 1 at 20:21
bjcolby15
7961616
asked Aug 1 at 17:54
RudY the Heck
42
Consider the following maximazation problem
Maximize $x^2+y-1$
Subject to $x^2+y^2<1$
Required...
1) Write Karush-Kuhn-Tucker equation that are needed to identify the extreme of $F$ on $A$
2) Determine the physical point that satisfy the Karush-Kuhn-Tucker's equation
economics
edited Aug 1 at 20:21
bjcolby15
7961616
asked Aug 1 at 17:54
RudY the Heck
42
edited Aug 1 at 20:21
bjcolby15
7961616
edited Aug 1 at 20:21
bjcolby15
7961616
edited Aug 1 at 20:21
bjcolby15
7961616
7961616
asked Aug 1 at 17:54
RudY the Heck
42
asked Aug 1 at 17:54
RudY the Heck
42
asked Aug 1 at 17:54
RudY the Heck
42
42
closed as off-topic by Isaac Browne, Ethan Bolker, mlc, Math Lover, callculus Aug 1 at 18:06
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Isaac Browne, Ethan Bolker, mlc, Math Lover, callculus
closed as off-topic by Isaac Browne, Ethan Bolker, mlc, Math Lover, callculus Aug 1 at 18:06
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Isaac Browne, Ethan Bolker, mlc, Math Lover, callculus
3
Kuhn tuckey, kuhn taken, kuhn takers...
– Kenny Lau
Aug 1 at 17:55
Kuhn turkey, tucker, ticker
– amWhy
Aug 1 at 18:04
Ok it's to say no idea rather than spittng such ......
– RudY the Heck
Aug 1 at 18:23
Follow this. The function $f(x,y)=x^2+y-1$. The function $g(x,y)=x^2+y^2-1$, which defines the restriction $g(x,y)leq0$ equivalent to $x^2+y^2leq1$. Notice that the restriction must have $leq$ and not $<$. Otherwise there is no maximum. Then $nabla f(x,y)=(2x,1)$, $nabla g(x,y)=(2x,2y)$. At an maximum $(x_0,y_0)$ we must have $(2x_0,1)=mu (2x_0,2y_0)$, with $x_0^2+y_0^2leq 1$, $mu leq 0$ and $mu(x_0^2+y_0^2-1)=0$.
– spiralstotheleft
Aug 2 at 2:03
add a comment |Â
3
Kuhn tuckey, kuhn taken, kuhn takers...
– Kenny Lau
Aug 1 at 17:55
Kuhn turkey, tucker, ticker
– amWhy
Aug 1 at 18:04
Ok it's to say no idea rather than spittng such ......
– RudY the Heck
Aug 1 at 18:23
Follow this. The function $f(x,y)=x^2+y-1$. The function $g(x,y)=x^2+y^2-1$, which defines the restriction $g(x,y)leq0$ equivalent to $x^2+y^2leq1$. Notice that the restriction must have $leq$ and not $<$. Otherwise there is no maximum. Then $nabla f(x,y)=(2x,1)$, $nabla g(x,y)=(2x,2y)$. At an maximum $(x_0,y_0)$ we must have $(2x_0,1)=mu (2x_0,2y_0)$, with $x_0^2+y_0^2leq 1$, $mu leq 0$ and $mu(x_0^2+y_0^2-1)=0$.
– spiralstotheleft
Aug 2 at 2:03
3
3
Kuhn tuckey, kuhn taken, kuhn takers...
– Kenny Lau
Aug 1 at 17:55
Kuhn tuckey, kuhn taken, kuhn takers...
– Kenny Lau
Aug 1 at 17:55
Kuhn turkey, tucker, ticker
– amWhy
Aug 1 at 18:04
Kuhn turkey, tucker, ticker
– amWhy
Aug 1 at 18:04
Ok it's to say no idea rather than spittng such ......
– RudY the Heck
Aug 1 at 18:23
Ok it's to say no idea rather than spittng such ......
– RudY the Heck
Aug 1 at 18:23
Follow this. The function $f(x,y)=x^2+y-1$. The function $g(x,y)=x^2+y^2-1$, which defines the restriction $g(x,y)leq0$ equivalent to $x^2+y^2leq1$. Notice that the restriction must have $leq$ and not $<$. Otherwise there is no maximum. Then $nabla f(x,y)=(2x,1)$, $nabla g(x,y)=(2x,2y)$. At an maximum $(x_0,y_0)$ we must have $(2x_0,1)=mu (2x_0,2y_0)$, with $x_0^2+y_0^2leq 1$, $mu leq 0$ and $mu(x_0^2+y_0^2-1)=0$.
– spiralstotheleft
Aug 2 at 2:03
Follow this. The function $f(x,y)=x^2+y-1$. The function $g(x,y)=x^2+y^2-1$, which defines the restriction $g(x,y)leq0$ equivalent to $x^2+y^2leq1$. Notice that the restriction must have $leq$ and not $<$. Otherwise there is no maximum. Then $nabla f(x,y)=(2x,1)$, $nabla g(x,y)=(2x,2y)$. At an maximum $(x_0,y_0)$ we must have $(2x_0,1)=mu (2x_0,2y_0)$, with $x_0^2+y_0^2leq 1$, $mu leq 0$ and $mu(x_0^2+y_0^2-1)=0$.
– spiralstotheleft
Aug 2 at 2:03
add a comment |Â
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3
Kuhn tuckey, kuhn taken, kuhn takers...
– Kenny Lau
Aug 1 at 17:55
Kuhn turkey, tucker, ticker
– amWhy
Aug 1 at 18:04
Ok it's to say no idea rather than spittng such ......
– RudY the Heck
Aug 1 at 18:23
Follow this. The function $f(x,y)=x^2+y-1$. The function $g(x,y)=x^2+y^2-1$, which defines the restriction $g(x,y)leq0$ equivalent to $x^2+y^2leq1$. Notice that the restriction must have $leq$ and not $<$. Otherwise there is no maximum. Then $nabla f(x,y)=(2x,1)$, $nabla g(x,y)=(2x,2y)$. At an maximum $(x_0,y_0)$ we must have $(2x_0,1)=mu (2x_0,2y_0)$, with $x_0^2+y_0^2leq 1$, $mu leq 0$ and $mu(x_0^2+y_0^2-1)=0$.
– spiralstotheleft
Aug 2 at 2:03