range of $3x^2-2xy$ subjected to $x^2+y^2=1$
$begingroup$
If $x^2+y^2=1$. then the range of expression $3x^2-2xy$ without trigonometric substitution method
what i have done try here is use arithmetic geometric inequality
$displaystyle x^2+y^2geq 2xy$
$displaystyle -2xygeq -(x^2+y^2)$
$displaystyle 3x^2-2xygeq 2x^2-y^2$
this will not help more
how do i solve it help me please
inequality
asked Dec 17 '18 at 18:28
jackyjacky
1,235815
$endgroup$
add a comment |
$begingroup$
If $x^2+y^2=1$. then the range of expression $3x^2-2xy$ without trigonometric substitution method
what i have done try here is use arithmetic geometric inequality
$displaystyle x^2+y^2geq 2xy$
$displaystyle -2xygeq -(x^2+y^2)$
$displaystyle 3x^2-2xygeq 2x^2-y^2$
this will not help more
how do i solve it help me please
inequality
asked Dec 17 '18 at 18:28
jackyjacky
1,235815
$endgroup$
$begingroup$
Maybe put $x=cos t, y=sin t$ and plug these into objective function, then constraint automatic and need to get range of resulting trig function.
$endgroup$
– coffeemath
Dec 17 '18 at 18:34
$begingroup$
"without trigonometric substitution"
$endgroup$
– David G. Stork
Dec 17 '18 at 18:39
add a comment |
$begingroup$
If $x^2+y^2=1$. then the range of expression $3x^2-2xy$ without trigonometric substitution method
what i have done try here is use arithmetic geometric inequality
$displaystyle x^2+y^2geq 2xy$
$displaystyle -2xygeq -(x^2+y^2)$
$displaystyle 3x^2-2xygeq 2x^2-y^2$
this will not help more
how do i solve it help me please
inequality
asked Dec 17 '18 at 18:28
jackyjacky
1,235815
$endgroup$
If $x^2+y^2=1$. then the range of expression $3x^2-2xy$ without trigonometric substitution method
what i have done try here is use arithmetic geometric inequality
$displaystyle x^2+y^2geq 2xy$
$displaystyle -2xygeq -(x^2+y^2)$
$displaystyle 3x^2-2xygeq 2x^2-y^2$
this will not help more
how do i solve it help me please
inequality
inequality
asked Dec 17 '18 at 18:28
jackyjacky
1,235815
asked Dec 17 '18 at 18:28
jackyjacky
1,235815
edited Dec 17 '18 at 18:31
jacky
asked Dec 17 '18 at 18:28
jackyjacky
1,235815
asked Dec 17 '18 at 18:28
jackyjacky
1,235815
asked Dec 17 '18 at 18:28
jackyjacky
1,235815
1,235815
$begingroup$
Maybe put $x=cos t, y=sin t$ and plug these into objective function, then constraint automatic and need to get range of resulting trig function.
$endgroup$
– coffeemath
Dec 17 '18 at 18:34
$begingroup$
"without trigonometric substitution"
$endgroup$
– David G. Stork
Dec 17 '18 at 18:39
add a comment |
$begingroup$
Maybe put $x=cos t, y=sin t$ and plug these into objective function, then constraint automatic and need to get range of resulting trig function.
$endgroup$
– coffeemath
Dec 17 '18 at 18:34
$begingroup$
"without trigonometric substitution"
$endgroup$
– David G. Stork
Dec 17 '18 at 18:39
$begingroup$
Maybe put $x=cos t, y=sin t$ and plug these into objective function, then constraint automatic and need to get range of resulting trig function.
$endgroup$
– coffeemath
Dec 17 '18 at 18:34
$begingroup$
Maybe put $x=cos t, y=sin t$ and plug these into objective function, then constraint automatic and need to get range of resulting trig function.
$endgroup$
– coffeemath
Dec 17 '18 at 18:34
$begingroup$
"without trigonometric substitution"
$endgroup$
– David G. Stork
Dec 17 '18 at 18:39
$begingroup$
"without trigonometric substitution"
$endgroup$
– David G. Stork
Dec 17 '18 at 18:39
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
If you have the requisite linear algebra theorems at your disposal, then you can recognize
$$3x^2-2xy=pmatrix{x&y}pmatrix{3&-1\-1&0}pmatrix{x\y}$$
where the $2times2$ matrix is selfadjoint with eigenvalues satisfying $lambda^2-3lambda-1=0$. For selfadjoint matrices $M$, the corresponding eigenvectors are orthogonal, and thus when the vector $mathbf{x}$ ranges over the unit ball (in this case the unit circle), $mathbf{x}^TMmathbf{x}$ ranges from the smallest eigenvalue to the largest, in this case, from $(3-sqrt{13})/2$ to $(3+sqrt{13})/2$.
answered Dec 17 '18 at 19:13
Barry CipraBarry Cipra
60.3k654127
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add a comment |
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Since $x^2+y^2=1$, you can substitute for $y=pmsqrt{1-x^2}$ in $f(x,y)=3x^2-2xy$ to get $g(x)=f(x,pmsqrt{1-x^2})=3x^2pm2xsqrt{1-x^2},-1le xle1$ and find the extrema using standard techniques.
answered Dec 17 '18 at 18:35
Shubham JohriShubham Johri
5,337718
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add a comment |
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The range that you're after is $left[frac{3-sqrt{13}}2,frac{3+sqrt{13}}2right]$. In fact, if $x^2+y^2=1$, then$$3x^2-2xy=3x^2pm xsqrt{1-x^2}.$$So, for each $xin[-1,1]$, let$$f(x)=3x^2-2xsqrt{1-x^2}text{ and let }g(x)=3x^2+xsqrt{1-x^2}.$$Now, using that standard Calculus techniques, you can check that the range of both functions $f$ and $g$ is $left[frac{3-sqrt{13}}2,frac{3+sqrt{13}}2right]$.
answered Dec 17 '18 at 18:36
José Carlos SantosJosé Carlos Santos
169k23132237
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add a comment |
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$3 x^2 - 2 x y$ with $y = pm sqrt{1-x^2}$ is $3 x^2 pm 2 x sqrt{1 - x^2}:$
Take derivative w.r.t. $x$, set it to zero and solve.
Note that the two functions are reflections about $x=0$, so their ranges are the same.
answered Dec 17 '18 at 18:34
David G. StorkDavid G. Stork
11.1k41432
$endgroup$
$begingroup$
Note $y$ could also be the negative of what you have.
$endgroup$
– coffeemath
Dec 17 '18 at 18:35
$begingroup$
It doesn't make a difference, the two functions are reflections of each other about $y$ axis and have the same range
$endgroup$
– Shubham Johri
Dec 17 '18 at 18:40
add a comment |
$begingroup$
For a calculs-free approach, note that
$$
frac{3+sqrt{13}}{2}left[frac{2y-(3+sqrt{13})x}{sqrt{26+6sqrt{13}}}right]^2+frac{3-sqrt{13}}{2}left[frac{2y-(3-sqrt{13})x}{sqrt{26-6sqrt{13}}}right]^2=3x^2-2xy
$$
and
$$
left[frac{2y-(3+sqrt{13})x}{sqrt{26+6sqrt{13}}}right]^2+left[frac{2y-(3-sqrt{13})x}{sqrt{26-6sqrt{13}}}right]^2=x^2+y^2
$$
So the range of $3x^2-2xy$, as $(x,y)$ varies on the unit circle, is the closed interval with endpoints $dfrac{3pmsqrt{13}}{2}$.
answered Dec 17 '18 at 18:49
user10354138user10354138
7,4322925
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add a comment |
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Making $y = lambda x$ and substituting we have
$$
mbox{Variation for} x^2(3-2lambda) mbox{s. t. } x^2(1+lambda^2) = 1
$$
or variation for
$$
f(lambda) = frac{3-2lambda}{1+lambda^2}
$$
now determining the stationary points with $f'(lambda) = 0$ obtaining
$$
lambda = frac 12left(3pmsqrt{13}right)
$$
etc.
answered Dec 17 '18 at 20:44
CesareoCesareo
9,4563517
$endgroup$
add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you have the requisite linear algebra theorems at your disposal, then you can recognize
$$3x^2-2xy=pmatrix{x&y}pmatrix{3&-1\-1&0}pmatrix{x\y}$$
where the $2times2$ matrix is selfadjoint with eigenvalues satisfying $lambda^2-3lambda-1=0$. For selfadjoint matrices $M$, the corresponding eigenvectors are orthogonal, and thus when the vector $mathbf{x}$ ranges over the unit ball (in this case the unit circle), $mathbf{x}^TMmathbf{x}$ ranges from the smallest eigenvalue to the largest, in this case, from $(3-sqrt{13})/2$ to $(3+sqrt{13})/2$.
answered Dec 17 '18 at 19:13
Barry CipraBarry Cipra
60.3k654127
$endgroup$
add a comment |
$begingroup$
If you have the requisite linear algebra theorems at your disposal, then you can recognize
$$3x^2-2xy=pmatrix{x&y}pmatrix{3&-1\-1&0}pmatrix{x\y}$$
where the $2times2$ matrix is selfadjoint with eigenvalues satisfying $lambda^2-3lambda-1=0$. For selfadjoint matrices $M$, the corresponding eigenvectors are orthogonal, and thus when the vector $mathbf{x}$ ranges over the unit ball (in this case the unit circle), $mathbf{x}^TMmathbf{x}$ ranges from the smallest eigenvalue to the largest, in this case, from $(3-sqrt{13})/2$ to $(3+sqrt{13})/2$.
answered Dec 17 '18 at 19:13
Barry CipraBarry Cipra
60.3k654127
$endgroup$
add a comment |
$begingroup$
If you have the requisite linear algebra theorems at your disposal, then you can recognize
$$3x^2-2xy=pmatrix{x&y}pmatrix{3&-1\-1&0}pmatrix{x\y}$$
where the $2times2$ matrix is selfadjoint with eigenvalues satisfying $lambda^2-3lambda-1=0$. For selfadjoint matrices $M$, the corresponding eigenvectors are orthogonal, and thus when the vector $mathbf{x}$ ranges over the unit ball (in this case the unit circle), $mathbf{x}^TMmathbf{x}$ ranges from the smallest eigenvalue to the largest, in this case, from $(3-sqrt{13})/2$ to $(3+sqrt{13})/2$.
answered Dec 17 '18 at 19:13
Barry CipraBarry Cipra
60.3k654127
$endgroup$
If you have the requisite linear algebra theorems at your disposal, then you can recognize
$$3x^2-2xy=pmatrix{x&y}pmatrix{3&-1\-1&0}pmatrix{x\y}$$
where the $2times2$ matrix is selfadjoint with eigenvalues satisfying $lambda^2-3lambda-1=0$. For selfadjoint matrices $M$, the corresponding eigenvectors are orthogonal, and thus when the vector $mathbf{x}$ ranges over the unit ball (in this case the unit circle), $mathbf{x}^TMmathbf{x}$ ranges from the smallest eigenvalue to the largest, in this case, from $(3-sqrt{13})/2$ to $(3+sqrt{13})/2$.
answered Dec 17 '18 at 19:13
Barry CipraBarry Cipra
60.3k654127
answered Dec 17 '18 at 19:13
Barry CipraBarry Cipra
60.3k654127
answered Dec 17 '18 at 19:13
Barry CipraBarry Cipra
60.3k654127
answered Dec 17 '18 at 19:13
Barry CipraBarry Cipra
60.3k654127
60.3k654127
add a comment |
add a comment |
$begingroup$
Since $x^2+y^2=1$, you can substitute for $y=pmsqrt{1-x^2}$ in $f(x,y)=3x^2-2xy$ to get $g(x)=f(x,pmsqrt{1-x^2})=3x^2pm2xsqrt{1-x^2},-1le xle1$ and find the extrema using standard techniques.
answered Dec 17 '18 at 18:35
Shubham JohriShubham Johri
5,337718
$endgroup$
add a comment |
$begingroup$
Since $x^2+y^2=1$, you can substitute for $y=pmsqrt{1-x^2}$ in $f(x,y)=3x^2-2xy$ to get $g(x)=f(x,pmsqrt{1-x^2})=3x^2pm2xsqrt{1-x^2},-1le xle1$ and find the extrema using standard techniques.
answered Dec 17 '18 at 18:35
Shubham JohriShubham Johri
5,337718
$endgroup$
add a comment |
$begingroup$
Since $x^2+y^2=1$, you can substitute for $y=pmsqrt{1-x^2}$ in $f(x,y)=3x^2-2xy$ to get $g(x)=f(x,pmsqrt{1-x^2})=3x^2pm2xsqrt{1-x^2},-1le xle1$ and find the extrema using standard techniques.
answered Dec 17 '18 at 18:35
Shubham JohriShubham Johri
5,337718
$endgroup$
Since $x^2+y^2=1$, you can substitute for $y=pmsqrt{1-x^2}$ in $f(x,y)=3x^2-2xy$ to get $g(x)=f(x,pmsqrt{1-x^2})=3x^2pm2xsqrt{1-x^2},-1le xle1$ and find the extrema using standard techniques.
answered Dec 17 '18 at 18:35
Shubham JohriShubham Johri
5,337718
answered Dec 17 '18 at 18:35
Shubham JohriShubham Johri
5,337718
answered Dec 17 '18 at 18:35
Shubham JohriShubham Johri
5,337718
answered Dec 17 '18 at 18:35
Shubham JohriShubham Johri
5,337718
5,337718
add a comment |
add a comment |
$begingroup$
The range that you're after is $left[frac{3-sqrt{13}}2,frac{3+sqrt{13}}2right]$. In fact, if $x^2+y^2=1$, then$$3x^2-2xy=3x^2pm xsqrt{1-x^2}.$$So, for each $xin[-1,1]$, let$$f(x)=3x^2-2xsqrt{1-x^2}text{ and let }g(x)=3x^2+xsqrt{1-x^2}.$$Now, using that standard Calculus techniques, you can check that the range of both functions $f$ and $g$ is $left[frac{3-sqrt{13}}2,frac{3+sqrt{13}}2right]$.
answered Dec 17 '18 at 18:36
José Carlos SantosJosé Carlos Santos
169k23132237
$endgroup$
add a comment |
$begingroup$
The range that you're after is $left[frac{3-sqrt{13}}2,frac{3+sqrt{13}}2right]$. In fact, if $x^2+y^2=1$, then$$3x^2-2xy=3x^2pm xsqrt{1-x^2}.$$So, for each $xin[-1,1]$, let$$f(x)=3x^2-2xsqrt{1-x^2}text{ and let }g(x)=3x^2+xsqrt{1-x^2}.$$Now, using that standard Calculus techniques, you can check that the range of both functions $f$ and $g$ is $left[frac{3-sqrt{13}}2,frac{3+sqrt{13}}2right]$.
answered Dec 17 '18 at 18:36
José Carlos SantosJosé Carlos Santos
169k23132237
$endgroup$
add a comment |
$begingroup$
The range that you're after is $left[frac{3-sqrt{13}}2,frac{3+sqrt{13}}2right]$. In fact, if $x^2+y^2=1$, then$$3x^2-2xy=3x^2pm xsqrt{1-x^2}.$$So, for each $xin[-1,1]$, let$$f(x)=3x^2-2xsqrt{1-x^2}text{ and let }g(x)=3x^2+xsqrt{1-x^2}.$$Now, using that standard Calculus techniques, you can check that the range of both functions $f$ and $g$ is $left[frac{3-sqrt{13}}2,frac{3+sqrt{13}}2right]$.
answered Dec 17 '18 at 18:36
José Carlos SantosJosé Carlos Santos
169k23132237
$endgroup$
The range that you're after is $left[frac{3-sqrt{13}}2,frac{3+sqrt{13}}2right]$. In fact, if $x^2+y^2=1$, then$$3x^2-2xy=3x^2pm xsqrt{1-x^2}.$$So, for each $xin[-1,1]$, let$$f(x)=3x^2-2xsqrt{1-x^2}text{ and let }g(x)=3x^2+xsqrt{1-x^2}.$$Now, using that standard Calculus techniques, you can check that the range of both functions $f$ and $g$ is $left[frac{3-sqrt{13}}2,frac{3+sqrt{13}}2right]$.
answered Dec 17 '18 at 18:36
José Carlos SantosJosé Carlos Santos
169k23132237
answered Dec 17 '18 at 18:36
José Carlos SantosJosé Carlos Santos
169k23132237
answered Dec 17 '18 at 18:36
José Carlos SantosJosé Carlos Santos
169k23132237
answered Dec 17 '18 at 18:36
José Carlos SantosJosé Carlos Santos
169k23132237
169k23132237
add a comment |
add a comment |
$begingroup$
$3 x^2 - 2 x y$ with $y = pm sqrt{1-x^2}$ is $3 x^2 pm 2 x sqrt{1 - x^2}:$
Take derivative w.r.t. $x$, set it to zero and solve.
Note that the two functions are reflections about $x=0$, so their ranges are the same.
answered Dec 17 '18 at 18:34
David G. StorkDavid G. Stork
11.1k41432
$endgroup$
$begingroup$
Note $y$ could also be the negative of what you have.
$endgroup$
– coffeemath
Dec 17 '18 at 18:35
$begingroup$
It doesn't make a difference, the two functions are reflections of each other about $y$ axis and have the same range
$endgroup$
– Shubham Johri
Dec 17 '18 at 18:40
add a comment |
$begingroup$
$3 x^2 - 2 x y$ with $y = pm sqrt{1-x^2}$ is $3 x^2 pm 2 x sqrt{1 - x^2}:$
Take derivative w.r.t. $x$, set it to zero and solve.
Note that the two functions are reflections about $x=0$, so their ranges are the same.
answered Dec 17 '18 at 18:34
David G. StorkDavid G. Stork
11.1k41432
$endgroup$
$begingroup$
Note $y$ could also be the negative of what you have.
$endgroup$
– coffeemath
Dec 17 '18 at 18:35
$begingroup$
It doesn't make a difference, the two functions are reflections of each other about $y$ axis and have the same range
$endgroup$
– Shubham Johri
Dec 17 '18 at 18:40
add a comment |
$begingroup$
$3 x^2 - 2 x y$ with $y = pm sqrt{1-x^2}$ is $3 x^2 pm 2 x sqrt{1 - x^2}:$
Take derivative w.r.t. $x$, set it to zero and solve.
Note that the two functions are reflections about $x=0$, so their ranges are the same.
answered Dec 17 '18 at 18:34
David G. StorkDavid G. Stork
11.1k41432
$endgroup$
$3 x^2 - 2 x y$ with $y = pm sqrt{1-x^2}$ is $3 x^2 pm 2 x sqrt{1 - x^2}:$
Take derivative w.r.t. $x$, set it to zero and solve.
Note that the two functions are reflections about $x=0$, so their ranges are the same.
answered Dec 17 '18 at 18:34
David G. StorkDavid G. Stork
11.1k41432
edited Dec 17 '18 at 18:44
answered Dec 17 '18 at 18:34
David G. StorkDavid G. Stork
11.1k41432
answered Dec 17 '18 at 18:34
David G. StorkDavid G. Stork
11.1k41432
answered Dec 17 '18 at 18:34
David G. StorkDavid G. Stork
11.1k41432
11.1k41432
$begingroup$
Note $y$ could also be the negative of what you have.
$endgroup$
– coffeemath
Dec 17 '18 at 18:35
$begingroup$
It doesn't make a difference, the two functions are reflections of each other about $y$ axis and have the same range
$endgroup$
– Shubham Johri
Dec 17 '18 at 18:40
add a comment |
$begingroup$
Note $y$ could also be the negative of what you have.
$endgroup$
– coffeemath
Dec 17 '18 at 18:35
$begingroup$
It doesn't make a difference, the two functions are reflections of each other about $y$ axis and have the same range
$endgroup$
– Shubham Johri
Dec 17 '18 at 18:40
$begingroup$
Note $y$ could also be the negative of what you have.
$endgroup$
– coffeemath
Dec 17 '18 at 18:35
$begingroup$
Note $y$ could also be the negative of what you have.
$endgroup$
– coffeemath
Dec 17 '18 at 18:35
$begingroup$
It doesn't make a difference, the two functions are reflections of each other about $y$ axis and have the same range
$endgroup$
– Shubham Johri
Dec 17 '18 at 18:40
$begingroup$
It doesn't make a difference, the two functions are reflections of each other about $y$ axis and have the same range
$endgroup$
– Shubham Johri
Dec 17 '18 at 18:40
add a comment |
$begingroup$
For a calculs-free approach, note that
$$
frac{3+sqrt{13}}{2}left[frac{2y-(3+sqrt{13})x}{sqrt{26+6sqrt{13}}}right]^2+frac{3-sqrt{13}}{2}left[frac{2y-(3-sqrt{13})x}{sqrt{26-6sqrt{13}}}right]^2=3x^2-2xy
$$
and
$$
left[frac{2y-(3+sqrt{13})x}{sqrt{26+6sqrt{13}}}right]^2+left[frac{2y-(3-sqrt{13})x}{sqrt{26-6sqrt{13}}}right]^2=x^2+y^2
$$
So the range of $3x^2-2xy$, as $(x,y)$ varies on the unit circle, is the closed interval with endpoints $dfrac{3pmsqrt{13}}{2}$.
answered Dec 17 '18 at 18:49
user10354138user10354138
7,4322925
$endgroup$
add a comment |
$begingroup$
For a calculs-free approach, note that
$$
frac{3+sqrt{13}}{2}left[frac{2y-(3+sqrt{13})x}{sqrt{26+6sqrt{13}}}right]^2+frac{3-sqrt{13}}{2}left[frac{2y-(3-sqrt{13})x}{sqrt{26-6sqrt{13}}}right]^2=3x^2-2xy
$$
and
$$
left[frac{2y-(3+sqrt{13})x}{sqrt{26+6sqrt{13}}}right]^2+left[frac{2y-(3-sqrt{13})x}{sqrt{26-6sqrt{13}}}right]^2=x^2+y^2
$$
So the range of $3x^2-2xy$, as $(x,y)$ varies on the unit circle, is the closed interval with endpoints $dfrac{3pmsqrt{13}}{2}$.
answered Dec 17 '18 at 18:49
user10354138user10354138
7,4322925
$endgroup$
add a comment |
$begingroup$
For a calculs-free approach, note that
$$
frac{3+sqrt{13}}{2}left[frac{2y-(3+sqrt{13})x}{sqrt{26+6sqrt{13}}}right]^2+frac{3-sqrt{13}}{2}left[frac{2y-(3-sqrt{13})x}{sqrt{26-6sqrt{13}}}right]^2=3x^2-2xy
$$
and
$$
left[frac{2y-(3+sqrt{13})x}{sqrt{26+6sqrt{13}}}right]^2+left[frac{2y-(3-sqrt{13})x}{sqrt{26-6sqrt{13}}}right]^2=x^2+y^2
$$
So the range of $3x^2-2xy$, as $(x,y)$ varies on the unit circle, is the closed interval with endpoints $dfrac{3pmsqrt{13}}{2}$.
answered Dec 17 '18 at 18:49
user10354138user10354138
7,4322925
$endgroup$
For a calculs-free approach, note that
$$
frac{3+sqrt{13}}{2}left[frac{2y-(3+sqrt{13})x}{sqrt{26+6sqrt{13}}}right]^2+frac{3-sqrt{13}}{2}left[frac{2y-(3-sqrt{13})x}{sqrt{26-6sqrt{13}}}right]^2=3x^2-2xy
$$
and
$$
left[frac{2y-(3+sqrt{13})x}{sqrt{26+6sqrt{13}}}right]^2+left[frac{2y-(3-sqrt{13})x}{sqrt{26-6sqrt{13}}}right]^2=x^2+y^2
$$
So the range of $3x^2-2xy$, as $(x,y)$ varies on the unit circle, is the closed interval with endpoints $dfrac{3pmsqrt{13}}{2}$.
answered Dec 17 '18 at 18:49
user10354138user10354138
7,4322925
answered Dec 17 '18 at 18:49
user10354138user10354138
7,4322925
answered Dec 17 '18 at 18:49
user10354138user10354138
7,4322925
answered Dec 17 '18 at 18:49
user10354138user10354138
7,4322925
7,4322925
add a comment |
add a comment |
$begingroup$
Making $y = lambda x$ and substituting we have
$$
mbox{Variation for} x^2(3-2lambda) mbox{s. t. } x^2(1+lambda^2) = 1
$$
or variation for
$$
f(lambda) = frac{3-2lambda}{1+lambda^2}
$$
now determining the stationary points with $f'(lambda) = 0$ obtaining
$$
lambda = frac 12left(3pmsqrt{13}right)
$$
etc.
answered Dec 17 '18 at 20:44
CesareoCesareo
9,4563517
$endgroup$
add a comment |
$begingroup$
Making $y = lambda x$ and substituting we have
$$
mbox{Variation for} x^2(3-2lambda) mbox{s. t. } x^2(1+lambda^2) = 1
$$
or variation for
$$
f(lambda) = frac{3-2lambda}{1+lambda^2}
$$
now determining the stationary points with $f'(lambda) = 0$ obtaining
$$
lambda = frac 12left(3pmsqrt{13}right)
$$
etc.
answered Dec 17 '18 at 20:44
CesareoCesareo
9,4563517
$endgroup$
add a comment |
$begingroup$
Making $y = lambda x$ and substituting we have
$$
mbox{Variation for} x^2(3-2lambda) mbox{s. t. } x^2(1+lambda^2) = 1
$$
or variation for
$$
f(lambda) = frac{3-2lambda}{1+lambda^2}
$$
now determining the stationary points with $f'(lambda) = 0$ obtaining
$$
lambda = frac 12left(3pmsqrt{13}right)
$$
etc.
answered Dec 17 '18 at 20:44
CesareoCesareo
9,4563517
$endgroup$
Making $y = lambda x$ and substituting we have
$$
mbox{Variation for} x^2(3-2lambda) mbox{s. t. } x^2(1+lambda^2) = 1
$$
or variation for
$$
f(lambda) = frac{3-2lambda}{1+lambda^2}
$$
now determining the stationary points with $f'(lambda) = 0$ obtaining
$$
lambda = frac 12left(3pmsqrt{13}right)
$$
etc.
answered Dec 17 '18 at 20:44
CesareoCesareo
9,4563517
answered Dec 17 '18 at 20:44
CesareoCesareo
9,4563517
answered Dec 17 '18 at 20:44
CesareoCesareo
9,4563517
answered Dec 17 '18 at 20:44
CesareoCesareo
9,4563517
9,4563517
add a comment |
add a comment |
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$begingroup$
Maybe put $x=cos t, y=sin t$ and plug these into objective function, then constraint automatic and need to get range of resulting trig function.
$endgroup$
– coffeemath
Dec 17 '18 at 18:34
$begingroup$
"without trigonometric substitution"
$endgroup$
– David G. Stork
Dec 17 '18 at 18:39