Use Lagrange multipliers to find all extrema of exponential function [answered]
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Using Lagrange multipliers, find the extreme values of $U(x,y)=e^{x+y}$ on the surface $x^2+xy+y^2=1$
$nabla U(x,y)=lambdanabla f(x,y) $ where $f(x,y)=x^2+xy+y^2-1$ (the constraint surface).
$$(e^{x+y},e^{x+y})=lambda(2x+y,2y+x)$$
Then
$$e^{x+y}=lambda(2x+y)$$
$$e^{x+y}=lambda(2y+x)$$
Therefore: $lambda >0$, $lambda(2x+y)=lambda(2y+x) Rightarrow 2x+y=2y+x Rightarrow x=y$
If $x=y$, then substituting in our constraint equation:
$x^2+x(x)+y^2=1 Rightarrow 3x^2=1 Rightarrow x=pmfrac{1}{sqrt{3}}$
Since, $x=y$, our extrema are $(frac{1}{sqrt{3}},frac{1}{sqrt{3}})$ and $(-frac{1}{sqrt{3}},-frac{1}{sqrt{3}})$
optimization lagrange-multiplier
asked Dec 11 '18 at 23:37
Kenny EKenny E
225
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add a comment |
$begingroup$
Using Lagrange multipliers, find the extreme values of $U(x,y)=e^{x+y}$ on the surface $x^2+xy+y^2=1$
$nabla U(x,y)=lambdanabla f(x,y) $ where $f(x,y)=x^2+xy+y^2-1$ (the constraint surface).
$$(e^{x+y},e^{x+y})=lambda(2x+y,2y+x)$$
Then
$$e^{x+y}=lambda(2x+y)$$
$$e^{x+y}=lambda(2y+x)$$
Therefore: $lambda >0$, $lambda(2x+y)=lambda(2y+x) Rightarrow 2x+y=2y+x Rightarrow x=y$
If $x=y$, then substituting in our constraint equation:
$x^2+x(x)+y^2=1 Rightarrow 3x^2=1 Rightarrow x=pmfrac{1}{sqrt{3}}$
Since, $x=y$, our extrema are $(frac{1}{sqrt{3}},frac{1}{sqrt{3}})$ and $(-frac{1}{sqrt{3}},-frac{1}{sqrt{3}})$
optimization lagrange-multiplier
asked Dec 11 '18 at 23:37
Kenny EKenny E
225
$endgroup$
add a comment |
$begingroup$
Using Lagrange multipliers, find the extreme values of $U(x,y)=e^{x+y}$ on the surface $x^2+xy+y^2=1$
$nabla U(x,y)=lambdanabla f(x,y) $ where $f(x,y)=x^2+xy+y^2-1$ (the constraint surface).
$$(e^{x+y},e^{x+y})=lambda(2x+y,2y+x)$$
Then
$$e^{x+y}=lambda(2x+y)$$
$$e^{x+y}=lambda(2y+x)$$
Therefore: $lambda >0$, $lambda(2x+y)=lambda(2y+x) Rightarrow 2x+y=2y+x Rightarrow x=y$
If $x=y$, then substituting in our constraint equation:
$x^2+x(x)+y^2=1 Rightarrow 3x^2=1 Rightarrow x=pmfrac{1}{sqrt{3}}$
Since, $x=y$, our extrema are $(frac{1}{sqrt{3}},frac{1}{sqrt{3}})$ and $(-frac{1}{sqrt{3}},-frac{1}{sqrt{3}})$
optimization lagrange-multiplier
asked Dec 11 '18 at 23:37
Kenny EKenny E
225
$endgroup$
Using Lagrange multipliers, find the extreme values of $U(x,y)=e^{x+y}$ on the surface $x^2+xy+y^2=1$
$nabla U(x,y)=lambdanabla f(x,y) $ where $f(x,y)=x^2+xy+y^2-1$ (the constraint surface).
$$(e^{x+y},e^{x+y})=lambda(2x+y,2y+x)$$
Then
$$e^{x+y}=lambda(2x+y)$$
$$e^{x+y}=lambda(2y+x)$$
Therefore: $lambda >0$, $lambda(2x+y)=lambda(2y+x) Rightarrow 2x+y=2y+x Rightarrow x=y$
If $x=y$, then substituting in our constraint equation:
$x^2+x(x)+y^2=1 Rightarrow 3x^2=1 Rightarrow x=pmfrac{1}{sqrt{3}}$
Since, $x=y$, our extrema are $(frac{1}{sqrt{3}},frac{1}{sqrt{3}})$ and $(-frac{1}{sqrt{3}},-frac{1}{sqrt{3}})$
optimization lagrange-multiplier
optimization lagrange-multiplier
asked Dec 11 '18 at 23:37
Kenny EKenny E
225
asked Dec 11 '18 at 23:37
Kenny EKenny E
225
edited Dec 13 '18 at 17:21
Kenny E
asked Dec 11 '18 at 23:37
Kenny EKenny E
225
asked Dec 11 '18 at 23:37
Kenny EKenny E
225
asked Dec 11 '18 at 23:37
Kenny EKenny E
225
225
add a comment |
add a comment |
1 Answer
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You immediately get $lambda>0$ and $2x+y=2y+x$, so $x=y$. Plugging this into the constraint, you get $3x^2=1$, so $x=pm 1/sqrt{3}$.
answered Dec 11 '18 at 23:41
LinAlgLinAlg
9,4761521
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Thank you! I had an epiphany shortly after posting, I'm going to add the rest of my solution now!
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– Kenny E
Dec 11 '18 at 23:52
1
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@KennyE you need $lambda>0$ to conclude $2x+y=2y+x$. Please mark this question as answered.
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– LinAlg
Dec 12 '18 at 14:30
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@KennyE you can mark this question as answered by checking the box next to my answer :)
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– LinAlg
Dec 24 '18 at 19:57
add a comment |
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1 Answer
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1 Answer
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You immediately get $lambda>0$ and $2x+y=2y+x$, so $x=y$. Plugging this into the constraint, you get $3x^2=1$, so $x=pm 1/sqrt{3}$.
answered Dec 11 '18 at 23:41
LinAlgLinAlg
9,4761521
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Thank you! I had an epiphany shortly after posting, I'm going to add the rest of my solution now!
$endgroup$
– Kenny E
Dec 11 '18 at 23:52
1
$begingroup$
@KennyE you need $lambda>0$ to conclude $2x+y=2y+x$. Please mark this question as answered.
$endgroup$
– LinAlg
Dec 12 '18 at 14:30
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@KennyE you can mark this question as answered by checking the box next to my answer :)
$endgroup$
– LinAlg
Dec 24 '18 at 19:57
add a comment |
$begingroup$
You immediately get $lambda>0$ and $2x+y=2y+x$, so $x=y$. Plugging this into the constraint, you get $3x^2=1$, so $x=pm 1/sqrt{3}$.
answered Dec 11 '18 at 23:41
LinAlgLinAlg
9,4761521
$endgroup$
$begingroup$
Thank you! I had an epiphany shortly after posting, I'm going to add the rest of my solution now!
$endgroup$
– Kenny E
Dec 11 '18 at 23:52
1
$begingroup$
@KennyE you need $lambda>0$ to conclude $2x+y=2y+x$. Please mark this question as answered.
$endgroup$
– LinAlg
Dec 12 '18 at 14:30
$begingroup$
@KennyE you can mark this question as answered by checking the box next to my answer :)
$endgroup$
– LinAlg
Dec 24 '18 at 19:57
add a comment |
$begingroup$
You immediately get $lambda>0$ and $2x+y=2y+x$, so $x=y$. Plugging this into the constraint, you get $3x^2=1$, so $x=pm 1/sqrt{3}$.
answered Dec 11 '18 at 23:41
LinAlgLinAlg
9,4761521
$endgroup$
You immediately get $lambda>0$ and $2x+y=2y+x$, so $x=y$. Plugging this into the constraint, you get $3x^2=1$, so $x=pm 1/sqrt{3}$.
answered Dec 11 '18 at 23:41
LinAlgLinAlg
9,4761521
edited Dec 12 '18 at 14:31
answered Dec 11 '18 at 23:41
LinAlgLinAlg
9,4761521
answered Dec 11 '18 at 23:41
LinAlgLinAlg
9,4761521
answered Dec 11 '18 at 23:41
LinAlgLinAlg
9,4761521
9,4761521
$begingroup$
Thank you! I had an epiphany shortly after posting, I'm going to add the rest of my solution now!
$endgroup$
– Kenny E
Dec 11 '18 at 23:52
1
$begingroup$
@KennyE you need $lambda>0$ to conclude $2x+y=2y+x$. Please mark this question as answered.
$endgroup$
– LinAlg
Dec 12 '18 at 14:30
$begingroup$
@KennyE you can mark this question as answered by checking the box next to my answer :)
$endgroup$
– LinAlg
Dec 24 '18 at 19:57
add a comment |
$begingroup$
Thank you! I had an epiphany shortly after posting, I'm going to add the rest of my solution now!
$endgroup$
– Kenny E
Dec 11 '18 at 23:52
1
$begingroup$
@KennyE you need $lambda>0$ to conclude $2x+y=2y+x$. Please mark this question as answered.
$endgroup$
– LinAlg
Dec 12 '18 at 14:30
$begingroup$
@KennyE you can mark this question as answered by checking the box next to my answer :)
$endgroup$
– LinAlg
Dec 24 '18 at 19:57
$begingroup$
Thank you! I had an epiphany shortly after posting, I'm going to add the rest of my solution now!
$endgroup$
– Kenny E
Dec 11 '18 at 23:52
$begingroup$
Thank you! I had an epiphany shortly after posting, I'm going to add the rest of my solution now!
$endgroup$
– Kenny E
Dec 11 '18 at 23:52
1
1
$begingroup$
@KennyE you need $lambda>0$ to conclude $2x+y=2y+x$. Please mark this question as answered.
$endgroup$
– LinAlg
Dec 12 '18 at 14:30
$begingroup$
@KennyE you need $lambda>0$ to conclude $2x+y=2y+x$. Please mark this question as answered.
$endgroup$
– LinAlg
Dec 12 '18 at 14:30
$begingroup$
@KennyE you can mark this question as answered by checking the box next to my answer :)
$endgroup$
– LinAlg
Dec 24 '18 at 19:57
$begingroup$
@KennyE you can mark this question as answered by checking the box next to my answer :)
$endgroup$
– LinAlg
Dec 24 '18 at 19:57
add a comment |
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