Mixing
Maximize a bivariate function under constraints by Lagrange multipliers
$begingroup$
I have been given a function $f(x,y)= x^2 + y^2$ whose maximum and minimum values have been sought (if existent) under the constraint that $3x^2 + 4xy +16y^2=140$.
This looks pretty much to be a problem, wherein the technique of Lagrange Multipliers can be exploited. So, I generate the Lagrangian and set the partial derivatives to 0; which gives the following set of equations:-
(i) $$2(1+3λ)x + 4λy = 0$$
(ii) $$4λx + 2(1+16λ)y = 0$$
(iii) The constraint itself; $$3x^2 + 4xy +16y^2=140$$.
But, solving this system is proving to be quite quite tedious.
Solving (i) and (ii); I retrieve $44λ^2+19λ+1=0$ whose solutions are not much-elegant (I know it's subjective!)
I then proceed to write $x$ in terms of $λ$ and $y$ from (i) and thereafter, substitute a value of $λ$ from above, getting $y$in terms of $x$.
Now, I substitute the above value of $y$ into (iii) which generates an equation in $x$; pending which the value of $x$ can be ascertained. Thereafter; the value of $y$ can be ascertained, too. And, then the substitution of these $x$ and $y$ into the parent equation of $f(x,y)= x^2 + y^2$ to get the desired result.
Now, I have to repeat the above two steps for the other value of $λ$, which just adds to the complexity.
So, my questions are:-
1) Does there exist any more efficient method for solving the system? I need to do this stuff at exam-hall where average time is around 8 mins and i assume that there might be a compulsion to use Lagrange Multipliers.
2) Can this sum be done in a more easy manner than using the concept of Lagrange multipliers; which may be employed in absence of the above compulsion?
multivariable-calculus lagrange-multiplier maxima-minima
asked Jan 12 at 13:50
Winged Blades of GodricWinged Blades of Godric
665
$endgroup$
add a comment |
$begingroup$
I have been given a function $f(x,y)= x^2 + y^2$ whose maximum and minimum values have been sought (if existent) under the constraint that $3x^2 + 4xy +16y^2=140$.
This looks pretty much to be a problem, wherein the technique of Lagrange Multipliers can be exploited. So, I generate the Lagrangian and set the partial derivatives to 0; which gives the following set of equations:-
(i) $$2(1+3λ)x + 4λy = 0$$
(ii) $$4λx + 2(1+16λ)y = 0$$
(iii) The constraint itself; $$3x^2 + 4xy +16y^2=140$$.
But, solving this system is proving to be quite quite tedious.
Solving (i) and (ii); I retrieve $44λ^2+19λ+1=0$ whose solutions are not much-elegant (I know it's subjective!)
I then proceed to write $x$ in terms of $λ$ and $y$ from (i) and thereafter, substitute a value of $λ$ from above, getting $y$in terms of $x$.
Now, I substitute the above value of $y$ into (iii) which generates an equation in $x$; pending which the value of $x$ can be ascertained. Thereafter; the value of $y$ can be ascertained, too. And, then the substitution of these $x$ and $y$ into the parent equation of $f(x,y)= x^2 + y^2$ to get the desired result.
Now, I have to repeat the above two steps for the other value of $λ$, which just adds to the complexity.
So, my questions are:-
1) Does there exist any more efficient method for solving the system? I need to do this stuff at exam-hall where average time is around 8 mins and i assume that there might be a compulsion to use Lagrange Multipliers.
2) Can this sum be done in a more easy manner than using the concept of Lagrange multipliers; which may be employed in absence of the above compulsion?
multivariable-calculus lagrange-multiplier maxima-minima
asked Jan 12 at 13:50
Winged Blades of GodricWinged Blades of Godric
665
$endgroup$
add a comment |
$begingroup$
I have been given a function $f(x,y)= x^2 + y^2$ whose maximum and minimum values have been sought (if existent) under the constraint that $3x^2 + 4xy +16y^2=140$.
This looks pretty much to be a problem, wherein the technique of Lagrange Multipliers can be exploited. So, I generate the Lagrangian and set the partial derivatives to 0; which gives the following set of equations:-
(i) $$2(1+3λ)x + 4λy = 0$$
(ii) $$4λx + 2(1+16λ)y = 0$$
(iii) The constraint itself; $$3x^2 + 4xy +16y^2=140$$.
But, solving this system is proving to be quite quite tedious.
Solving (i) and (ii); I retrieve $44λ^2+19λ+1=0$ whose solutions are not much-elegant (I know it's subjective!)
I then proceed to write $x$ in terms of $λ$ and $y$ from (i) and thereafter, substitute a value of $λ$ from above, getting $y$in terms of $x$.
Now, I substitute the above value of $y$ into (iii) which generates an equation in $x$; pending which the value of $x$ can be ascertained. Thereafter; the value of $y$ can be ascertained, too. And, then the substitution of these $x$ and $y$ into the parent equation of $f(x,y)= x^2 + y^2$ to get the desired result.
Now, I have to repeat the above two steps for the other value of $λ$, which just adds to the complexity.
So, my questions are:-
1) Does there exist any more efficient method for solving the system? I need to do this stuff at exam-hall where average time is around 8 mins and i assume that there might be a compulsion to use Lagrange Multipliers.
2) Can this sum be done in a more easy manner than using the concept of Lagrange multipliers; which may be employed in absence of the above compulsion?
multivariable-calculus lagrange-multiplier maxima-minima
asked Jan 12 at 13:50
Winged Blades of GodricWinged Blades of Godric
665
$endgroup$
I have been given a function $f(x,y)= x^2 + y^2$ whose maximum and minimum values have been sought (if existent) under the constraint that $3x^2 + 4xy +16y^2=140$.
This looks pretty much to be a problem, wherein the technique of Lagrange Multipliers can be exploited. So, I generate the Lagrangian and set the partial derivatives to 0; which gives the following set of equations:-
(i) $$2(1+3λ)x + 4λy = 0$$
(ii) $$4λx + 2(1+16λ)y = 0$$
(iii) The constraint itself; $$3x^2 + 4xy +16y^2=140$$.
But, solving this system is proving to be quite quite tedious.
Solving (i) and (ii); I retrieve $44λ^2+19λ+1=0$ whose solutions are not much-elegant (I know it's subjective!)
I then proceed to write $x$ in terms of $λ$ and $y$ from (i) and thereafter, substitute a value of $λ$ from above, getting $y$in terms of $x$.
Now, I substitute the above value of $y$ into (iii) which generates an equation in $x$; pending which the value of $x$ can be ascertained. Thereafter; the value of $y$ can be ascertained, too. And, then the substitution of these $x$ and $y$ into the parent equation of $f(x,y)= x^2 + y^2$ to get the desired result.
Now, I have to repeat the above two steps for the other value of $λ$, which just adds to the complexity.
So, my questions are:-
1) Does there exist any more efficient method for solving the system? I need to do this stuff at exam-hall where average time is around 8 mins and i assume that there might be a compulsion to use Lagrange Multipliers.
2) Can this sum be done in a more easy manner than using the concept of Lagrange multipliers; which may be employed in absence of the above compulsion?
multivariable-calculus lagrange-multiplier maxima-minima
multivariable-calculus lagrange-multiplier maxima-minima
asked Jan 12 at 13:50
Winged Blades of GodricWinged Blades of Godric
665
asked Jan 12 at 13:50
Winged Blades of GodricWinged Blades of Godric
665
edited Jan 12 at 14:11
Winged Blades of Godric
asked Jan 12 at 13:50
Winged Blades of GodricWinged Blades of Godric
665
asked Jan 12 at 13:50
Winged Blades of GodricWinged Blades of Godric
665
asked Jan 12 at 13:50
Winged Blades of GodricWinged Blades of Godric
665
665
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Here's one way: Substitute $x=rcos theta$ and $y=rsintheta$. Then, the constraint becomes
$$
3cos^2theta +4sinthetacostheta +16sin^2theta = frac{140}{r^2}.
$$Using the double angle formula and $sin^2 theta+cos^2 theta= 1$, this is equivalent to
$$
2sin2theta -frac{13}{2}cos2theta = frac{140}{r^2}-frac{19}{2}.
$$ The LHS takes values in $[-sqrt{2^2+(frac{13}{2})^2},sqrt{2^2+(frac{13}{2})^2}]$ so the minimum and maximum of $r^2$ can be obtained by solving
$$
frac{140}{r^2}-frac{19}{2}=pmfrac{sqrt{185}}{2}.
$$ This gives $$
r^2=frac{35(19pmsqrt{185})}{22}.
$$
answered Jan 12 at 13:58
SongSong
12.4k630
$endgroup$
$begingroup$
This's an excellent solution :-) Many thanks!
$endgroup$
– Winged Blades of Godric
Jan 12 at 14:56
$begingroup$
I hope this will help :)
$endgroup$
– Song
Jan 12 at 15:17
add a comment |
$begingroup$
If you examine the geometry of this particular problem, you might find that using a Lagrange multiplier is entirely unnecessary. The objective function is the square of the distance from the origin and the constraint is a conic of some sort, so what you are being asked to do is to find the square of the maximum and minimum distances of the conic from the origin. These are just the squares of the semiaxis lengths, which are the reciprocals of the eigenvalues of the associated matrix: $$frac1{140}begin{bmatrix}3&2\2&16end{bmatrix}.$$ The problem therefore reduces to solving the characteristic equation $lambda^2-19lambda+44=0$.
answered Jan 12 at 21:59
amdamd
30k21050
$endgroup$
add a comment |
$begingroup$
Solution without Lagrange multipliers. Let $x^2+y^2=k$.
Thus, the equation $$x^2+y^2=frac{k(3x^2+4xy+16y^2)}{140}$$ or
$$(3k-140)x^2+4kxy+(16k-140)y^2=0$$ has solutions, which for $kneqfrac{140}{3}$ gives
$$4k^2-(3k-140)(16k-140)geq0$$ or
$$frac{665-35sqrt{185}}{22}leq kleq frac{665+35sqrt{185}}{22}.$$
We see that $$frac{665-35sqrt{185}}{22}leq frac{140}{3}leq frac{665+35sqrt{185}}{22}$$ and since the equality occurs for $$x=-frac{2ky}{3k-140},$$ we got the maximalk value and the minimal value.
answered Jan 12 at 14:31
Michael RozenbergMichael Rozenberg
103k1891195
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here's one way: Substitute $x=rcos theta$ and $y=rsintheta$. Then, the constraint becomes
$$
3cos^2theta +4sinthetacostheta +16sin^2theta = frac{140}{r^2}.
$$Using the double angle formula and $sin^2 theta+cos^2 theta= 1$, this is equivalent to
$$
2sin2theta -frac{13}{2}cos2theta = frac{140}{r^2}-frac{19}{2}.
$$ The LHS takes values in $[-sqrt{2^2+(frac{13}{2})^2},sqrt{2^2+(frac{13}{2})^2}]$ so the minimum and maximum of $r^2$ can be obtained by solving
$$
frac{140}{r^2}-frac{19}{2}=pmfrac{sqrt{185}}{2}.
$$ This gives $$
r^2=frac{35(19pmsqrt{185})}{22}.
$$
answered Jan 12 at 13:58
SongSong
12.4k630
$endgroup$
$begingroup$
This's an excellent solution :-) Many thanks!
$endgroup$
– Winged Blades of Godric
Jan 12 at 14:56
$begingroup$
I hope this will help :)
$endgroup$
– Song
Jan 12 at 15:17
add a comment |
$begingroup$
Here's one way: Substitute $x=rcos theta$ and $y=rsintheta$. Then, the constraint becomes
$$
3cos^2theta +4sinthetacostheta +16sin^2theta = frac{140}{r^2}.
$$Using the double angle formula and $sin^2 theta+cos^2 theta= 1$, this is equivalent to
$$
2sin2theta -frac{13}{2}cos2theta = frac{140}{r^2}-frac{19}{2}.
$$ The LHS takes values in $[-sqrt{2^2+(frac{13}{2})^2},sqrt{2^2+(frac{13}{2})^2}]$ so the minimum and maximum of $r^2$ can be obtained by solving
$$
frac{140}{r^2}-frac{19}{2}=pmfrac{sqrt{185}}{2}.
$$ This gives $$
r^2=frac{35(19pmsqrt{185})}{22}.
$$
answered Jan 12 at 13:58
SongSong
12.4k630
$endgroup$
$begingroup$
This's an excellent solution :-) Many thanks!
$endgroup$
– Winged Blades of Godric
Jan 12 at 14:56
$begingroup$
I hope this will help :)
$endgroup$
– Song
Jan 12 at 15:17
add a comment |
$begingroup$
Here's one way: Substitute $x=rcos theta$ and $y=rsintheta$. Then, the constraint becomes
$$
3cos^2theta +4sinthetacostheta +16sin^2theta = frac{140}{r^2}.
$$Using the double angle formula and $sin^2 theta+cos^2 theta= 1$, this is equivalent to
$$
2sin2theta -frac{13}{2}cos2theta = frac{140}{r^2}-frac{19}{2}.
$$ The LHS takes values in $[-sqrt{2^2+(frac{13}{2})^2},sqrt{2^2+(frac{13}{2})^2}]$ so the minimum and maximum of $r^2$ can be obtained by solving
$$
frac{140}{r^2}-frac{19}{2}=pmfrac{sqrt{185}}{2}.
$$ This gives $$
r^2=frac{35(19pmsqrt{185})}{22}.
$$
answered Jan 12 at 13:58
SongSong
12.4k630
$endgroup$
Here's one way: Substitute $x=rcos theta$ and $y=rsintheta$. Then, the constraint becomes
$$
3cos^2theta +4sinthetacostheta +16sin^2theta = frac{140}{r^2}.
$$Using the double angle formula and $sin^2 theta+cos^2 theta= 1$, this is equivalent to
$$
2sin2theta -frac{13}{2}cos2theta = frac{140}{r^2}-frac{19}{2}.
$$ The LHS takes values in $[-sqrt{2^2+(frac{13}{2})^2},sqrt{2^2+(frac{13}{2})^2}]$ so the minimum and maximum of $r^2$ can be obtained by solving
$$
frac{140}{r^2}-frac{19}{2}=pmfrac{sqrt{185}}{2}.
$$ This gives $$
r^2=frac{35(19pmsqrt{185})}{22}.
$$
answered Jan 12 at 13:58
SongSong
12.4k630
edited Jan 13 at 2:07
answered Jan 12 at 13:58
SongSong
12.4k630
answered Jan 12 at 13:58
SongSong
12.4k630
answered Jan 12 at 13:58
SongSong
12.4k630
12.4k630
$begingroup$
This's an excellent solution :-) Many thanks!
$endgroup$
– Winged Blades of Godric
Jan 12 at 14:56
$begingroup$
I hope this will help :)
$endgroup$
– Song
Jan 12 at 15:17
add a comment |
$begingroup$
This's an excellent solution :-) Many thanks!
$endgroup$
– Winged Blades of Godric
Jan 12 at 14:56
$begingroup$
I hope this will help :)
$endgroup$
– Song
Jan 12 at 15:17
$begingroup$
This's an excellent solution :-) Many thanks!
$endgroup$
– Winged Blades of Godric
Jan 12 at 14:56
$begingroup$
This's an excellent solution :-) Many thanks!
$endgroup$
– Winged Blades of Godric
Jan 12 at 14:56
$begingroup$
I hope this will help :)
$endgroup$
– Song
Jan 12 at 15:17
$begingroup$
I hope this will help :)
$endgroup$
– Song
Jan 12 at 15:17
add a comment |
$begingroup$
If you examine the geometry of this particular problem, you might find that using a Lagrange multiplier is entirely unnecessary. The objective function is the square of the distance from the origin and the constraint is a conic of some sort, so what you are being asked to do is to find the square of the maximum and minimum distances of the conic from the origin. These are just the squares of the semiaxis lengths, which are the reciprocals of the eigenvalues of the associated matrix: $$frac1{140}begin{bmatrix}3&2\2&16end{bmatrix}.$$ The problem therefore reduces to solving the characteristic equation $lambda^2-19lambda+44=0$.
answered Jan 12 at 21:59
amdamd
30k21050
$endgroup$
add a comment |
$begingroup$
If you examine the geometry of this particular problem, you might find that using a Lagrange multiplier is entirely unnecessary. The objective function is the square of the distance from the origin and the constraint is a conic of some sort, so what you are being asked to do is to find the square of the maximum and minimum distances of the conic from the origin. These are just the squares of the semiaxis lengths, which are the reciprocals of the eigenvalues of the associated matrix: $$frac1{140}begin{bmatrix}3&2\2&16end{bmatrix}.$$ The problem therefore reduces to solving the characteristic equation $lambda^2-19lambda+44=0$.
answered Jan 12 at 21:59
amdamd
30k21050
$endgroup$
add a comment |
$begingroup$
If you examine the geometry of this particular problem, you might find that using a Lagrange multiplier is entirely unnecessary. The objective function is the square of the distance from the origin and the constraint is a conic of some sort, so what you are being asked to do is to find the square of the maximum and minimum distances of the conic from the origin. These are just the squares of the semiaxis lengths, which are the reciprocals of the eigenvalues of the associated matrix: $$frac1{140}begin{bmatrix}3&2\2&16end{bmatrix}.$$ The problem therefore reduces to solving the characteristic equation $lambda^2-19lambda+44=0$.
answered Jan 12 at 21:59
amdamd
30k21050
$endgroup$
If you examine the geometry of this particular problem, you might find that using a Lagrange multiplier is entirely unnecessary. The objective function is the square of the distance from the origin and the constraint is a conic of some sort, so what you are being asked to do is to find the square of the maximum and minimum distances of the conic from the origin. These are just the squares of the semiaxis lengths, which are the reciprocals of the eigenvalues of the associated matrix: $$frac1{140}begin{bmatrix}3&2\2&16end{bmatrix}.$$ The problem therefore reduces to solving the characteristic equation $lambda^2-19lambda+44=0$.
answered Jan 12 at 21:59
amdamd
30k21050
answered Jan 12 at 21:59
amdamd
30k21050
answered Jan 12 at 21:59
amdamd
30k21050
answered Jan 12 at 21:59
amdamd
30k21050
30k21050
add a comment |
add a comment |
$begingroup$
Solution without Lagrange multipliers. Let $x^2+y^2=k$.
Thus, the equation $$x^2+y^2=frac{k(3x^2+4xy+16y^2)}{140}$$ or
$$(3k-140)x^2+4kxy+(16k-140)y^2=0$$ has solutions, which for $kneqfrac{140}{3}$ gives
$$4k^2-(3k-140)(16k-140)geq0$$ or
$$frac{665-35sqrt{185}}{22}leq kleq frac{665+35sqrt{185}}{22}.$$
We see that $$frac{665-35sqrt{185}}{22}leq frac{140}{3}leq frac{665+35sqrt{185}}{22}$$ and since the equality occurs for $$x=-frac{2ky}{3k-140},$$ we got the maximalk value and the minimal value.
answered Jan 12 at 14:31
Michael RozenbergMichael Rozenberg
103k1891195
$endgroup$
add a comment |
$begingroup$
Solution without Lagrange multipliers. Let $x^2+y^2=k$.
Thus, the equation $$x^2+y^2=frac{k(3x^2+4xy+16y^2)}{140}$$ or
$$(3k-140)x^2+4kxy+(16k-140)y^2=0$$ has solutions, which for $kneqfrac{140}{3}$ gives
$$4k^2-(3k-140)(16k-140)geq0$$ or
$$frac{665-35sqrt{185}}{22}leq kleq frac{665+35sqrt{185}}{22}.$$
We see that $$frac{665-35sqrt{185}}{22}leq frac{140}{3}leq frac{665+35sqrt{185}}{22}$$ and since the equality occurs for $$x=-frac{2ky}{3k-140},$$ we got the maximalk value and the minimal value.
answered Jan 12 at 14:31
Michael RozenbergMichael Rozenberg
103k1891195
$endgroup$
add a comment |
$begingroup$
Solution without Lagrange multipliers. Let $x^2+y^2=k$.
Thus, the equation $$x^2+y^2=frac{k(3x^2+4xy+16y^2)}{140}$$ or
$$(3k-140)x^2+4kxy+(16k-140)y^2=0$$ has solutions, which for $kneqfrac{140}{3}$ gives
$$4k^2-(3k-140)(16k-140)geq0$$ or
$$frac{665-35sqrt{185}}{22}leq kleq frac{665+35sqrt{185}}{22}.$$
We see that $$frac{665-35sqrt{185}}{22}leq frac{140}{3}leq frac{665+35sqrt{185}}{22}$$ and since the equality occurs for $$x=-frac{2ky}{3k-140},$$ we got the maximalk value and the minimal value.
answered Jan 12 at 14:31
Michael RozenbergMichael Rozenberg
103k1891195
$endgroup$
Solution without Lagrange multipliers. Let $x^2+y^2=k$.
Thus, the equation $$x^2+y^2=frac{k(3x^2+4xy+16y^2)}{140}$$ or
$$(3k-140)x^2+4kxy+(16k-140)y^2=0$$ has solutions, which for $kneqfrac{140}{3}$ gives
$$4k^2-(3k-140)(16k-140)geq0$$ or
$$frac{665-35sqrt{185}}{22}leq kleq frac{665+35sqrt{185}}{22}.$$
We see that $$frac{665-35sqrt{185}}{22}leq frac{140}{3}leq frac{665+35sqrt{185}}{22}$$ and since the equality occurs for $$x=-frac{2ky}{3k-140},$$ we got the maximalk value and the minimal value.
answered Jan 12 at 14:31
Michael RozenbergMichael Rozenberg
103k1891195
answered Jan 12 at 14:31
Michael RozenbergMichael Rozenberg
103k1891195
answered Jan 12 at 14:31
Michael RozenbergMichael Rozenberg
103k1891195
answered Jan 12 at 14:31
Michael RozenbergMichael Rozenberg
103k1891195
103k1891195
add a comment |
add a comment |
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