MHB GWR309's question at Yahoo Answers regarding Lagrange multipliers

[MarkFL]

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=1Additional 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.

 

[MarkFL]

Hello GWR309,

We are given the objective function:

$$f(x,y)=xy$$

subject to the constraint:

$$g(x,y)=x-2y-1=0$$

Using Lagrange multipliers, we get the system:

$$y=\lambda$$

$$x=-2\lambda$$

and this implies:

$$x=-2y$$

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

$$-2y-2y-1=0$$

$$y=-\frac{1}{4}\,\therefore\,x=\frac{1}{2}$$

And so the critical point is:

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

and the optimize value for the objective function is:

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

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

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

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

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

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:

$$x=\frac{1}{2}$$

and so the minimum is:

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

To GWR309 and any other guests viewing this topic, I invite and encourage you to post other optimization problems in our http://www.mathhelpboards.com/f10/ forum.

Best Regards,

Mark.

 

[GWR309]

Hello GWR309,

We are given the objective function:

$$f(x,y)=xy$$

subject to the constraint:

$$g(x,y)=x-2y-1=0$$

Using Lagrange multipliers, we get the system:

$$y=\lambda$$

$$x=-2\lambda$$

and this implies:

$$x=-2y$$

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

$$-2y-2y-1=0$$

$$y=-\frac{1}{4}\,\therefore\,x=\frac{1}{2}$$

And so the critical point is:

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

and the optimize value for the objective function is:

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

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

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

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

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

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:

$$x=\frac{1}{2}$$

and so the minimum is:

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

To GWR309 and any other guests viewing this topic, I invite and encourage you to post other optimization problems in our http://www.mathhelpboards.com/f10/ forum.

Best Regards,

Mark.

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

x=−2λ

 

[MarkFL]

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:

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

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

$$g(x,y)=0$$

Among the solutions $(x,y,\lambda)$ of the system will be the points $$\left(x_i,y_i \right)$$, 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 $$f\left(x_i,y_i \right)$$.

So, given that:

$$f(x,y)=xy$$

$$g(x,y)=x-2y-1=0$$

we then find by computing the first partials, that:

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

and so the system we are to solve is:

$$y=\lambda$$

$$x=-2\lambda$$

$$g(x,y)=x-2y-1=0$$

From the first two equations, we find:

$$\lambda=y=-\frac{x}{2}\,\therefore\,x=-2y$$

and the rest follows as in my second post above.

 

[GWR309]

for general lagrange multipliers problems, how do I determine if it is a max, min or neither?

 

[Prove It]

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...