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Lagrange multipliers for extreme values

skatenerd

Active member
Oct 3, 2012
114
The problem given is to find the local extreme values of \(f(x,y)=x^2y\) on the line \(x+y=3\). I went through the system of equations with the partial derivatives of \(x\), \(y\), and \(\lambda\), and found two extreme points \((0,3)\) and \((2,1)\). Plugging that into the original function I found \(f(0,3)=0\) and \(f(2,1)=4\). So from what I have learned so far, that means that \((0,3)\) is the minimum and \((2,1)\) is the maximum, correct?
Then how come if I find some other point that lies on the line \(x+y=3\) such as \((8,-5)\), that would be a negative number when plugged into \(x^2y\)? Wouldn't mean that \((0,3)\) isn't actually the minimum? Maybe I am misunderstanding the way these work, but if somebody could explain this to me that would be nice. Thanks
 

chisigma

Well-known member
Feb 13, 2012
1,704
Re: lagrange multipliers for extreme values

The problem given is to find the local extreme values of \(f(x,y)=x^2y\) on the line \(x+y=3\). I went through the system of equations with the partial derivatives of \(x\), \(y\), and \(\lambda\), and found two extreme points \((0,3)\) and \((2,1)\). Plugging that into the original function I found \(f(0,3)=0\) and \(f(2,1)=4\). So from what I have learned so far, that means that \((0,3)\) is the minimum and \((2,1)\) is the maximum, correct?
Then how come if I find some other point that lies on the line \(x+y=3\) such as \((8,-5)\), that would be a negative number when plugged into \(x^2y\)? Wouldn't mean that \((0,3)\) isn't actually the minimum? Maybe I am misunderstanding the way these work, but if somebody could explain this to me that would be nice. Thanks
In this case You don't need to use Lagrange multipliers because the problem is to find the maxima and minima of the function $\displaystyle f(x)= 3\ x^{2} - x^{3}$, that has a local maximum in $x=2$ and a local minimum in $x=0$. However $\displaystyle f(x)= 3\ x^{2} - x^{3}$ tends to $- \infty$ if x tends to $\infty$ and vice versa so that it has neither and absolte maximum nor an absolute minimum...


Kind regards

$\chi$ $\sigma$
 

skatenerd

Active member
Oct 3, 2012
114
Re: lagrange multipliers for extreme values

Wait so are you saying that with these situations with one constraint to the original function you can just find the intersection of the two? If that is the case then what is the point of ever using lagrange multipliers when you have only one constraint?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: lagrange multipliers for extreme values

It is possible to express the objective function in one variable in this case because the constraint may be explicitly solved for one or both variables. It has been my experience that using Lagrange multipliers is computationally simpler, even in such cases.