Welcome to our community

Be a part of something great, join today!

GWR309's question at Yahoo! Answers regarding Lagrange multipliers

  • Thread starter
  • Admin
  • #1

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Here is the question:

Please help with lagrange multipliers?

I don't understand how to do these at all. Here is one of the problems I have to do:

f(x,y) = xy x-2y=1


Additional Details i thought I had more spaces when I typed... xy and x-2y=1 are 2 different equations.
Here is a link to the question:

Please help with lagrange multipliers? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
  • Thread starter
  • Admin
  • #2

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Hello GWR309,

We are given the objective function:

\(\displaystyle f(x,y)=xy\)

subject to the constraint:

\(\displaystyle g(x,y)=x-2y-1=0\)

Using Lagrange multipliers, we get the system:

\(\displaystyle y=\lambda\)

\(\displaystyle x=-2\lambda\)

and this implies:

\(\displaystyle x=-2y\)

Substituting for $x$ into the constraint, we find:

\(\displaystyle -2y-2y-1=0\)

\(\displaystyle y=-\frac{1}{4}\,\therefore\,x=\frac{1}{2}\)

And so the critical point is:

\(\displaystyle (x,y)=\left(\frac{1}{2},-\frac{1}{4} \right)\)

and the optimize value for the objective function is:

\(\displaystyle f\left(\frac{1}{2},-\frac{1}{4} \right)=-\frac{1}{8}\)

Since other test points such as \(\displaystyle \left(0,-\frac{1}{2} \right)\) and \(\displaystyle \left(1,0 \right)\) give \(\displaystyle f(x,y)>-\frac{1}{8}\) we may conclude that:

\(\displaystyle f_{\min}=f\left(\frac{1}{2},-\frac{1}{4} \right)=-\frac{1}{8}\)

As a check, we may use the constraint to write $f$ as a function of 1 variable, namely:

\(\displaystyle f(x)=\frac{x}{2}(x-1)\)

We can see now that $f(x)$ is a parabola opening upwards, and we know its vertex (which is the global minimum) will line on the axis of symmetry, which is midway between the roots, on the line:

\(\displaystyle x=\frac{1}{2}\)

and so the minimum is:

\(\displaystyle f_{\min}=f\left(\frac{1}{2} \right)=-\frac{1}{8}\)

To GWR309 and any other guests viewing this topic, I invite and encourage you to post other optimization problems in our Calculus forum.

Best Regards,

Mark.
 

GWR309

New member
May 8, 2013
4
Hello GWR309,

We are given the objective function:

\(\displaystyle f(x,y)=xy\)

subject to the constraint:

\(\displaystyle g(x,y)=x-2y-1=0\)

Using Lagrange multipliers, we get the system:

\(\displaystyle y=\lambda\)

\(\displaystyle x=-2\lambda\)

and this implies:

\(\displaystyle x=-2y\)

Substituting for $x$ into the constraint, we find:

\(\displaystyle -2y-2y-1=0\)

\(\displaystyle y=-\frac{1}{4}\,\therefore\,x=\frac{1}{2}\)

And so the critical point is:

\(\displaystyle (x,y)=\left(\frac{1}{2},-\frac{1}{4} \right)\)

and the optimize value for the objective function is:

\(\displaystyle f\left(\frac{1}{2},-\frac{1}{4} \right)=-\frac{1}{8}\)

Since other test points such as \(\displaystyle \left(0,-\frac{1}{2} \right)\) and \(\displaystyle \left(1,0 \right)\) give \(\displaystyle f(x,y)>-\frac{1}{8}\) we may conclude that:

\(\displaystyle f_{\min}=f\left(\frac{1}{2},-\frac{1}{4} \right)=-\frac{1}{8}\)

As a check, we may use the constraint to write $f$ as a function of 1 variable, namely:

\(\displaystyle f(x)=\frac{x}{2}(x-1)\)

We can see now that $f(x)$ is a parabola opening upwards, and we know its vertex (which is the global minimum) will line on the axis of symmetry, which is midway between the roots, on the line:

\(\displaystyle x=\frac{1}{2}\)

and so the minimum is:

\(\displaystyle f_{\min}=f\left(\frac{1}{2} \right)=-\frac{1}{8}\)

To GWR309 and any other guests viewing this topic, I invite and encourage you to post other optimization problems in our Calculus forum.

Best Regards,

Mark.
Where did you get the system though? this:
y=λ

x=−2λ
 
  • Thread starter
  • Admin
  • #4

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Where did you get the system though? this:
y=λ

x=−2λ
Hello GWR309,

Glad you joined us here! (Cool)

The method of Lagrange multipliers states:

To find the extrema of $z=f(x,y)$ subject to the constraint $g(x,y)=0$, solve the system of equations:

\(\displaystyle f_x(x,y)=\lambda g_x(x,y)\)

\(\displaystyle f_y(x,y)=\lambda g_y(x,y)\)

\(\displaystyle g(x,y)=0\)

Among the solutions $(x,y,\lambda)$ of the system will be the points \(\displaystyle \left(x_i,y_i \right)\), where $f$ has an extremum. When $f$ has a maximum (or minimum), it will be the largest (or smallest) number in the list of functional values \(\displaystyle f\left(x_i,y_i \right)\).

So, given that:

\(\displaystyle f(x,y)=xy\)

\(\displaystyle g(x,y)=x-2y-1=0\)

we then find by computing the first partials, that:

\(\displaystyle f_x(x,y)=y,\,f_y(x,y)=x,\,g_x(x,y)=1,\,g_y(x,y)=-2\)

and so the system we are to solve is:

\(\displaystyle y=\lambda\)

\(\displaystyle x=-2\lambda\)

\(\displaystyle g(x,y)=x-2y-1=0\)

From the first two equations, we find:

\(\displaystyle \lambda=y=-\frac{x}{2}\,\therefore\,x=-2y\)

and the rest follows as in my second post above.
 

GWR309

New member
May 8, 2013
4
for general lagrange multipliers problems, how do I determine if it is a max, min or neither?
 
Last edited:

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
for general lagrange multipliers problems, how do I determine if it is a max, min or neither?
You could check the Hessian matrix. Or check that your function is convex (which would imply you have a minimum). Or even just substitute in the value and compare it to the endpoints of your function...