# GWR309's question at Yahoo! Answers regarding Lagrange multipliers

#### MarkFL

Staff member
Here is the question:

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:

I have posted a link there to this topic so the OP can find my response.

#### MarkFL

Staff member
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
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

#### MarkFL

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

x=−2λ
Hello GWR309,

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
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
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...