# Using Lagrange multipliers to find max and min values?

First off, I'd like to apologize, but I'm unfamiliar with how to format math equations on this platform and am not finding anything to fix that.

I've been given the problem: Use Lagrange multipliers to find the maximum and minimum values of $f$(if they exist) subject to the given constraint.

$f(x,y) = 6xy$ subject to $x^2+y^2 -xy = 9$

I know that the components of $\nabla f(x,y)$ are equal to the components of $\nabla g(x,y)$ times the Lagrange multiplier, but when I set up a system of equations I am at a loss.

# l = Lagrange multiplier

$6y = l(2x-y)$, $6x = l(2y-x)$, $x^2+y^2-xy-9=0$

None of the examples in the book explain how to deal with instances in which components of the gradient have both x and y.

I don't know how to tackle this system of equations. Could somebody point me in the right direction?

• This was a little trickier than I thought. The first thing one needs to do is to establish that a minimum and maximum exist. I have done this and will write it up, but I suspect I am missing an easier approach. Commented Oct 25, 2015 at 22:00

You need to check if a minimum and maximum exist first.

A standard way is to check that the constraints define a compact set. It is clear that the feasible set is closed, so we need only establish boundedness.

Suppose $(x,y) \neq (0,0)$, then dividing the constraint by $x^2+y^2$ gives $1 = {9 \over x^2+y^2} + {x y \over x^2+y^2}$. Since $|xy| \le {1 \over 2} (x^2+y^2)$, we get ${9 \over x^2+y^2} \ge {1 \over 2}$ which shows that the feasible set is bounded.

So, we know that a maximiser and a minimiser exists.

Solving the equations is not too difficult. You have $6y+\lambda(2x-y) = 0$, $6x+\lambda(2y-x) = 0$ which you can write as $y(6-\lambda) = -2 \lambda x$ and $x(6-\lambda) = -2 \lambda y$.

We see that $x=0$ iff $y=0$. So suppose $x \neq 0$ (and hence $y \neq 0$). Dividing the equations gives $x^2 = y^2$ (justify why $\lambda \notin \{0,6\}$).

Hence all solutions to these equations are of the form $(x,x)$ or $(x,-x)$.

If we look for solutions of the form $(x,x)$, the constraint gives $2x^2 -x^2 = 9$, or $x = \pm 3$.

If we look for solutions of the form $(x,-x)$, the constraint gives $2x^2 +x^2 = 9$, or $x = \pm {1 \over \sqrt{3}}$.

Hence we have only four candidates to check.