I am working on a problem asking me to use the method of constrained extrema to find the global maximum and minimum for the following functions:
$$f(x,y) = xy$$ constrained to $$g(x,y) = 4x^2 + 2xy + y^2 - 4$$
So far I have done the calculations and solved the problem and have come up with an answer that does not make sense however I cannot see where I may have made a calculation error.
The first thing I did was find the gradient of both f(x,y) and g(x,y) and equate them to get the following result:
$$ y = 8x\lambda + 2y\lambda$$ $$ x = 2x\lambda + 2y\lambda$$
I multiplied the first equation by x and the second equation by y then divided both by lambda to get the following:
$$ \frac{xy}{\lambda} = 8x^2 + 2xy$$ $$ \frac{xy}{\lambda} = 2xy + 2y^2$$
As the left hand side of both equations are equal I produced the following: $$ 8x^2 + 2xy = 2xy + 2y^2$$ I then simplified this and did the algebra to come up with a solution y = 2x which I then substituted back into g(x,y) and got the solution $$ x = \pm \frac{2}{\sqrt{12}}$$
And I believe this result is incorrect as when you plug these values into f(x,y) you get the same answer of one third for both values. This must be incorrect as one of these values should be a maximum and the other a minimum as asked in the question.
Thank you,
Michael
Your setup is correct, in as much as you do indeed want to solve the system of three equations $$\nabla f = \lambda \nabla g$$ $$g = 0.$$
However you have to be extremely careful when doing so, as generally you will have many local extrema and saddle points in addition to the global extrema you search for. It is very easy to accidentally drop possible solutions.
For instance, in your first step, eliminating $\lambda$ from the first two equations is a fine idea. However you have implicitly assumed $\lambda \neq 0$; you must check that there isn't a solution with $\lambda =0$ to the set of equations above (there is not, though there could have been if $g(0,0)=0$.)
Next you have the system
$$ \begin{align*} 4x^2 &= y^2\\ 4x^2+2xy+y^2-4&=0. \end{align*} $$
You didn't explain how you "did some algebra" but you missed two solutions. I would approach the above by first looking at the first equation, which gives me $y = \pm 2x$; substituting each into the bottom equation gives me a quadratic in $x$, so I expect to get up to four solutions. In this case all four solutions are real and must be tested for global optimality.
Finally: a global maximum and minimum may not exist (if the feasible set is noncompact) or may not be attained at critical points (if the feasible set has a boundary, which arises in inequality-constrained problem; or worse, if your objective/constraint are not smooth). For this problem, the last thing to check for a thorough solution is whether $g=0$ is bounded. This can be seen by e.g. inspecting the Hessian of $g$ and noticing that it is positive-definite.