Mixing
Max and min of $f(x,y)=e^x+e^y$ on $x^2+y^2=1$
$begingroup$
$V={(x,y):x^2+y^2=1}$ is compact and f is continuous on V so for Weierstrass global max and min exist.To find them I have used lagrange multipliers but the system don't give a clear solution.
real-analysis
asked Jan 31 at 9:32
user495707user495707
25
$endgroup$
add a comment |
$begingroup$
$V={(x,y):x^2+y^2=1}$ is compact and f is continuous on V so for Weierstrass global max and min exist.To find them I have used lagrange multipliers but the system don't give a clear solution.
real-analysis
asked Jan 31 at 9:32
user495707user495707
25
$endgroup$
$begingroup$
I have a wild shot... Substitute $y$ by $pmsqrt{1-x^2}$ and put it in $f$. Then apply normal differentiation process to find maxima and minima. I don't know what this will result, just a vague idea to tackle such problems.
$endgroup$
– Anik Bhowmick
Jan 31 at 9:42
$begingroup$
One can find the minima in a completely elementary manner - by AM-GM, $e^x + e^y ge 2e^{(x+y)/2}$, with equality along the line $x = y.$ But trivially the minimiser of $(x+y)$ on the unit circle also lies on this line, and immediately we have that the minima of the original objective is $2exp(-1/sqrt{2}).$ Question: can the maxima be arrived at through some similar elementary method? I can't quite think of one.
$endgroup$
– stochasticboy321
Jan 31 at 9:58
add a comment |
$begingroup$
$V={(x,y):x^2+y^2=1}$ is compact and f is continuous on V so for Weierstrass global max and min exist.To find them I have used lagrange multipliers but the system don't give a clear solution.
real-analysis
asked Jan 31 at 9:32
user495707user495707
25
$endgroup$
$V={(x,y):x^2+y^2=1}$ is compact and f is continuous on V so for Weierstrass global max and min exist.To find them I have used lagrange multipliers but the system don't give a clear solution.
real-analysis
real-analysis
asked Jan 31 at 9:32
user495707user495707
25
asked Jan 31 at 9:32
user495707user495707
25
asked Jan 31 at 9:32
user495707user495707
25
asked Jan 31 at 9:32
user495707user495707
25
asked Jan 31 at 9:32
user495707user495707
25
25
$begingroup$
I have a wild shot... Substitute $y$ by $pmsqrt{1-x^2}$ and put it in $f$. Then apply normal differentiation process to find maxima and minima. I don't know what this will result, just a vague idea to tackle such problems.
$endgroup$
– Anik Bhowmick
Jan 31 at 9:42
$begingroup$
One can find the minima in a completely elementary manner - by AM-GM, $e^x + e^y ge 2e^{(x+y)/2}$, with equality along the line $x = y.$ But trivially the minimiser of $(x+y)$ on the unit circle also lies on this line, and immediately we have that the minima of the original objective is $2exp(-1/sqrt{2}).$ Question: can the maxima be arrived at through some similar elementary method? I can't quite think of one.
$endgroup$
– stochasticboy321
Jan 31 at 9:58
add a comment |
$begingroup$
I have a wild shot... Substitute $y$ by $pmsqrt{1-x^2}$ and put it in $f$. Then apply normal differentiation process to find maxima and minima. I don't know what this will result, just a vague idea to tackle such problems.
$endgroup$
– Anik Bhowmick
Jan 31 at 9:42
$begingroup$
One can find the minima in a completely elementary manner - by AM-GM, $e^x + e^y ge 2e^{(x+y)/2}$, with equality along the line $x = y.$ But trivially the minimiser of $(x+y)$ on the unit circle also lies on this line, and immediately we have that the minima of the original objective is $2exp(-1/sqrt{2}).$ Question: can the maxima be arrived at through some similar elementary method? I can't quite think of one.
$endgroup$
– stochasticboy321
Jan 31 at 9:58
$begingroup$
I have a wild shot... Substitute $y$ by $pmsqrt{1-x^2}$ and put it in $f$. Then apply normal differentiation process to find maxima and minima. I don't know what this will result, just a vague idea to tackle such problems.
$endgroup$
– Anik Bhowmick
Jan 31 at 9:42
$begingroup$
I have a wild shot... Substitute $y$ by $pmsqrt{1-x^2}$ and put it in $f$. Then apply normal differentiation process to find maxima and minima. I don't know what this will result, just a vague idea to tackle such problems.
$endgroup$
– Anik Bhowmick
Jan 31 at 9:42
$begingroup$
One can find the minima in a completely elementary manner - by AM-GM, $e^x + e^y ge 2e^{(x+y)/2}$, with equality along the line $x = y.$ But trivially the minimiser of $(x+y)$ on the unit circle also lies on this line, and immediately we have that the minima of the original objective is $2exp(-1/sqrt{2}).$ Question: can the maxima be arrived at through some similar elementary method? I can't quite think of one.
$endgroup$
– stochasticboy321
Jan 31 at 9:58
$begingroup$
One can find the minima in a completely elementary manner - by AM-GM, $e^x + e^y ge 2e^{(x+y)/2}$, with equality along the line $x = y.$ But trivially the minimiser of $(x+y)$ on the unit circle also lies on this line, and immediately we have that the minima of the original objective is $2exp(-1/sqrt{2}).$ Question: can the maxima be arrived at through some similar elementary method? I can't quite think of one.
$endgroup$
– stochasticboy321
Jan 31 at 9:58
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The contour curves $y=ln (z-e^x)$ and the constraint circle $x^2+y^2=1$ are shown on the graph:
$hspace{1cm}$
The min/max of $z(x,y)$ will occur when $x=y$.
answered Jan 31 at 9:48
farruhotafarruhota
21.8k2842
$endgroup$
add a comment |
$begingroup$
To find the maximum it is enough to consider positive values of $x$ and $y$ because $(x,y) in V$ implies $(pm x ,pm y) in V$. The method of Lagrange multipliers leads to the equation $xe^{y}=ye^{x}$. Note that the derivative of $frac {e^{x}} x$ is negative on $(0,1)$ so we get $x=y$. Hence the maximum is attained at $x=y=frac 1 {sqrt 2}$. Hint for minimum: the minimum value will be attained when $x,y<0$.
answered Jan 31 at 9:44
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
$endgroup$
add a comment |
$begingroup$
No need to use Lagrange at all. Substitute $x$ and $y$ by $sin(t)$ and $cos(t)$ the problem becomes maximizing $g(t):=( e^{sin(t)} + e^{cos(t)} )$. Then $g'(t) = e^{sin(t)}cos(t) - e^{cos(t)} sin(t) $. Then $g'(t)$ then is zero iff
$$ frac{cos(t)}{ e^{cos(t)}} = frac{sin(t)}{ e^{sin(t)}} $$
Let $ h(t) = t e^{-t} $. Then the upper equation becomes $h(cos(t)) = h(sin(t))$ . As $ h $ is monotonic increasing(as $h'ge 0$ on [0,1]) the equation will only be true where $cos(t) = sin(t)$. So the equation will only be true for $t = pi/4, text{ or } 5pi/4 $, or $sin(t) =cos(t) = pm sqrt{2}/2 $. So Max will be attained as $2e^{sqrt{2}/2}$ and min as $2e^{-sqrt{2}/2}$
answered Jan 31 at 10:24
MaksimMaksim
1,00719
$endgroup$
1
$begingroup$
It is not true for $t = - pi/4 $. $sin(- pi/4)={- sqrt{2}/2}$ and $cos(- pi/4)={sqrt{2}/2}$
$endgroup$
– Piotr Wasilewicz
Feb 1 at 10:02
1
$begingroup$
Good catch! Corrected.
$endgroup$
– Maksim
Feb 1 at 10:30
1
$begingroup$
I think you mean $5pi/4$ :)
$endgroup$
– Piotr Wasilewicz
Feb 1 at 11:21
$begingroup$
You are so right :-)
$endgroup$
– Maksim
Feb 1 at 11:53
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The contour curves $y=ln (z-e^x)$ and the constraint circle $x^2+y^2=1$ are shown on the graph:
$hspace{1cm}$
The min/max of $z(x,y)$ will occur when $x=y$.
answered Jan 31 at 9:48
farruhotafarruhota
21.8k2842
$endgroup$
add a comment |
$begingroup$
The contour curves $y=ln (z-e^x)$ and the constraint circle $x^2+y^2=1$ are shown on the graph:
$hspace{1cm}$
The min/max of $z(x,y)$ will occur when $x=y$.
answered Jan 31 at 9:48
farruhotafarruhota
21.8k2842
$endgroup$
add a comment |
$begingroup$
The contour curves $y=ln (z-e^x)$ and the constraint circle $x^2+y^2=1$ are shown on the graph:
$hspace{1cm}$
The min/max of $z(x,y)$ will occur when $x=y$.
answered Jan 31 at 9:48
farruhotafarruhota
21.8k2842
$endgroup$
The contour curves $y=ln (z-e^x)$ and the constraint circle $x^2+y^2=1$ are shown on the graph:
$hspace{1cm}$
The min/max of $z(x,y)$ will occur when $x=y$.
answered Jan 31 at 9:48
farruhotafarruhota
21.8k2842
answered Jan 31 at 9:48
farruhotafarruhota
21.8k2842
answered Jan 31 at 9:48
farruhotafarruhota
21.8k2842
answered Jan 31 at 9:48
farruhotafarruhota
21.8k2842
21.8k2842
add a comment |
add a comment |
$begingroup$
To find the maximum it is enough to consider positive values of $x$ and $y$ because $(x,y) in V$ implies $(pm x ,pm y) in V$. The method of Lagrange multipliers leads to the equation $xe^{y}=ye^{x}$. Note that the derivative of $frac {e^{x}} x$ is negative on $(0,1)$ so we get $x=y$. Hence the maximum is attained at $x=y=frac 1 {sqrt 2}$. Hint for minimum: the minimum value will be attained when $x,y<0$.
answered Jan 31 at 9:44
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
$endgroup$
add a comment |
$begingroup$
To find the maximum it is enough to consider positive values of $x$ and $y$ because $(x,y) in V$ implies $(pm x ,pm y) in V$. The method of Lagrange multipliers leads to the equation $xe^{y}=ye^{x}$. Note that the derivative of $frac {e^{x}} x$ is negative on $(0,1)$ so we get $x=y$. Hence the maximum is attained at $x=y=frac 1 {sqrt 2}$. Hint for minimum: the minimum value will be attained when $x,y<0$.
answered Jan 31 at 9:44
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
$endgroup$
add a comment |
$begingroup$
To find the maximum it is enough to consider positive values of $x$ and $y$ because $(x,y) in V$ implies $(pm x ,pm y) in V$. The method of Lagrange multipliers leads to the equation $xe^{y}=ye^{x}$. Note that the derivative of $frac {e^{x}} x$ is negative on $(0,1)$ so we get $x=y$. Hence the maximum is attained at $x=y=frac 1 {sqrt 2}$. Hint for minimum: the minimum value will be attained when $x,y<0$.
answered Jan 31 at 9:44
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
$endgroup$
To find the maximum it is enough to consider positive values of $x$ and $y$ because $(x,y) in V$ implies $(pm x ,pm y) in V$. The method of Lagrange multipliers leads to the equation $xe^{y}=ye^{x}$. Note that the derivative of $frac {e^{x}} x$ is negative on $(0,1)$ so we get $x=y$. Hence the maximum is attained at $x=y=frac 1 {sqrt 2}$. Hint for minimum: the minimum value will be attained when $x,y<0$.
answered Jan 31 at 9:44
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
edited Feb 1 at 11:41
answered Jan 31 at 9:44
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
answered Jan 31 at 9:44
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
answered Jan 31 at 9:44
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
72.6k53170
add a comment |
add a comment |
$begingroup$
No need to use Lagrange at all. Substitute $x$ and $y$ by $sin(t)$ and $cos(t)$ the problem becomes maximizing $g(t):=( e^{sin(t)} + e^{cos(t)} )$. Then $g'(t) = e^{sin(t)}cos(t) - e^{cos(t)} sin(t) $. Then $g'(t)$ then is zero iff
$$ frac{cos(t)}{ e^{cos(t)}} = frac{sin(t)}{ e^{sin(t)}} $$
Let $ h(t) = t e^{-t} $. Then the upper equation becomes $h(cos(t)) = h(sin(t))$ . As $ h $ is monotonic increasing(as $h'ge 0$ on [0,1]) the equation will only be true where $cos(t) = sin(t)$. So the equation will only be true for $t = pi/4, text{ or } 5pi/4 $, or $sin(t) =cos(t) = pm sqrt{2}/2 $. So Max will be attained as $2e^{sqrt{2}/2}$ and min as $2e^{-sqrt{2}/2}$
answered Jan 31 at 10:24
MaksimMaksim
1,00719
$endgroup$
1
$begingroup$
It is not true for $t = - pi/4 $. $sin(- pi/4)={- sqrt{2}/2}$ and $cos(- pi/4)={sqrt{2}/2}$
$endgroup$
– Piotr Wasilewicz
Feb 1 at 10:02
1
$begingroup$
Good catch! Corrected.
$endgroup$
– Maksim
Feb 1 at 10:30
1
$begingroup$
I think you mean $5pi/4$ :)
$endgroup$
– Piotr Wasilewicz
Feb 1 at 11:21
$begingroup$
You are so right :-)
$endgroup$
– Maksim
Feb 1 at 11:53
add a comment |
$begingroup$
No need to use Lagrange at all. Substitute $x$ and $y$ by $sin(t)$ and $cos(t)$ the problem becomes maximizing $g(t):=( e^{sin(t)} + e^{cos(t)} )$. Then $g'(t) = e^{sin(t)}cos(t) - e^{cos(t)} sin(t) $. Then $g'(t)$ then is zero iff
$$ frac{cos(t)}{ e^{cos(t)}} = frac{sin(t)}{ e^{sin(t)}} $$
Let $ h(t) = t e^{-t} $. Then the upper equation becomes $h(cos(t)) = h(sin(t))$ . As $ h $ is monotonic increasing(as $h'ge 0$ on [0,1]) the equation will only be true where $cos(t) = sin(t)$. So the equation will only be true for $t = pi/4, text{ or } 5pi/4 $, or $sin(t) =cos(t) = pm sqrt{2}/2 $. So Max will be attained as $2e^{sqrt{2}/2}$ and min as $2e^{-sqrt{2}/2}$
answered Jan 31 at 10:24
MaksimMaksim
1,00719
$endgroup$
1
$begingroup$
It is not true for $t = - pi/4 $. $sin(- pi/4)={- sqrt{2}/2}$ and $cos(- pi/4)={sqrt{2}/2}$
$endgroup$
– Piotr Wasilewicz
Feb 1 at 10:02
1
$begingroup$
Good catch! Corrected.
$endgroup$
– Maksim
Feb 1 at 10:30
1
$begingroup$
I think you mean $5pi/4$ :)
$endgroup$
– Piotr Wasilewicz
Feb 1 at 11:21
$begingroup$
You are so right :-)
$endgroup$
– Maksim
Feb 1 at 11:53
add a comment |
$begingroup$
No need to use Lagrange at all. Substitute $x$ and $y$ by $sin(t)$ and $cos(t)$ the problem becomes maximizing $g(t):=( e^{sin(t)} + e^{cos(t)} )$. Then $g'(t) = e^{sin(t)}cos(t) - e^{cos(t)} sin(t) $. Then $g'(t)$ then is zero iff
$$ frac{cos(t)}{ e^{cos(t)}} = frac{sin(t)}{ e^{sin(t)}} $$
Let $ h(t) = t e^{-t} $. Then the upper equation becomes $h(cos(t)) = h(sin(t))$ . As $ h $ is monotonic increasing(as $h'ge 0$ on [0,1]) the equation will only be true where $cos(t) = sin(t)$. So the equation will only be true for $t = pi/4, text{ or } 5pi/4 $, or $sin(t) =cos(t) = pm sqrt{2}/2 $. So Max will be attained as $2e^{sqrt{2}/2}$ and min as $2e^{-sqrt{2}/2}$
answered Jan 31 at 10:24
MaksimMaksim
1,00719
$endgroup$
No need to use Lagrange at all. Substitute $x$ and $y$ by $sin(t)$ and $cos(t)$ the problem becomes maximizing $g(t):=( e^{sin(t)} + e^{cos(t)} )$. Then $g'(t) = e^{sin(t)}cos(t) - e^{cos(t)} sin(t) $. Then $g'(t)$ then is zero iff
$$ frac{cos(t)}{ e^{cos(t)}} = frac{sin(t)}{ e^{sin(t)}} $$
Let $ h(t) = t e^{-t} $. Then the upper equation becomes $h(cos(t)) = h(sin(t))$ . As $ h $ is monotonic increasing(as $h'ge 0$ on [0,1]) the equation will only be true where $cos(t) = sin(t)$. So the equation will only be true for $t = pi/4, text{ or } 5pi/4 $, or $sin(t) =cos(t) = pm sqrt{2}/2 $. So Max will be attained as $2e^{sqrt{2}/2}$ and min as $2e^{-sqrt{2}/2}$
answered Jan 31 at 10:24
MaksimMaksim
1,00719
edited Feb 1 at 11:53
answered Jan 31 at 10:24
MaksimMaksim
1,00719
answered Jan 31 at 10:24
MaksimMaksim
1,00719
answered Jan 31 at 10:24
MaksimMaksim
1,00719
1,00719
1
$begingroup$
It is not true for $t = - pi/4 $. $sin(- pi/4)={- sqrt{2}/2}$ and $cos(- pi/4)={sqrt{2}/2}$
$endgroup$
– Piotr Wasilewicz
Feb 1 at 10:02
1
$begingroup$
Good catch! Corrected.
$endgroup$
– Maksim
Feb 1 at 10:30
1
$begingroup$
I think you mean $5pi/4$ :)
$endgroup$
– Piotr Wasilewicz
Feb 1 at 11:21
$begingroup$
You are so right :-)
$endgroup$
– Maksim
Feb 1 at 11:53
add a comment |
1
$begingroup$
It is not true for $t = - pi/4 $. $sin(- pi/4)={- sqrt{2}/2}$ and $cos(- pi/4)={sqrt{2}/2}$
$endgroup$
– Piotr Wasilewicz
Feb 1 at 10:02
1
$begingroup$
Good catch! Corrected.
$endgroup$
– Maksim
Feb 1 at 10:30
1
$begingroup$
I think you mean $5pi/4$ :)
$endgroup$
– Piotr Wasilewicz
Feb 1 at 11:21
$begingroup$
You are so right :-)
$endgroup$
– Maksim
Feb 1 at 11:53
1
1
$begingroup$
It is not true for $t = - pi/4 $. $sin(- pi/4)={- sqrt{2}/2}$ and $cos(- pi/4)={sqrt{2}/2}$
$endgroup$
– Piotr Wasilewicz
Feb 1 at 10:02
$begingroup$
It is not true for $t = - pi/4 $. $sin(- pi/4)={- sqrt{2}/2}$ and $cos(- pi/4)={sqrt{2}/2}$
$endgroup$
– Piotr Wasilewicz
Feb 1 at 10:02
1
1
$begingroup$
Good catch! Corrected.
$endgroup$
– Maksim
Feb 1 at 10:30
$begingroup$
Good catch! Corrected.
$endgroup$
– Maksim
Feb 1 at 10:30
1
1
$begingroup$
I think you mean $5pi/4$ :)
$endgroup$
– Piotr Wasilewicz
Feb 1 at 11:21
$begingroup$
I think you mean $5pi/4$ :)
$endgroup$
– Piotr Wasilewicz
Feb 1 at 11:21
$begingroup$
You are so right :-)
$endgroup$
– Maksim
Feb 1 at 11:53
$begingroup$
You are so right :-)
$endgroup$
– Maksim
Feb 1 at 11:53
add a comment |
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$begingroup$
I have a wild shot... Substitute $y$ by $pmsqrt{1-x^2}$ and put it in $f$. Then apply normal differentiation process to find maxima and minima. I don't know what this will result, just a vague idea to tackle such problems.
$endgroup$
– Anik Bhowmick
Jan 31 at 9:42
$begingroup$
One can find the minima in a completely elementary manner - by AM-GM, $e^x + e^y ge 2e^{(x+y)/2}$, with equality along the line $x = y.$ But trivially the minimiser of $(x+y)$ on the unit circle also lies on this line, and immediately we have that the minima of the original objective is $2exp(-1/sqrt{2}).$ Question: can the maxima be arrived at through some similar elementary method? I can't quite think of one.
$endgroup$
– stochasticboy321
Jan 31 at 9:58