Using Lagrange Multipliers with Constraints of a Line and a Parabola Find the absolute maximum and minimum of $f(x,y)= y^2-2xy+x^3-x$ on the region bounded by the curve $y=x^2$ and the line $y=4$. You must use Lagrange Multipliers to study the function on the curve $y=x^2$.
I'm unsure how to approach this because $y=4$ is given. Is this a trick question?
 A: You need to look for points within the region and on the boundary of the region which will be candidates for the absolute maximum and minimum of $f$. In the interior of the region, you should look for critical points by finding what points $P$ give $\nabla f(P) = 0$. However, you also have to check the values of $f$ on the boundary. In this case the boundary above will be the line $y=4$, while the boundary below will be $y=x^2$. You need to check both boundaries! One method to check for the possible candidates for maxima and minima which are on the boundary is to use the method of Lagrange multipliers. For example, if you want to check for extreme points of $f$ on $y=x^2$, you can set $g(x,y) = y-x^2$ and apply the Lagrange multiplier method giving you the system of equations
$$
y=x^2 \\
\nabla f = \lambda \nabla g
$$
Make sure to pay attention to the fact that the boundary $y=x^2$ only considers points with $-2 \leq x \leq 2$, since otherwise you've gone past your other boundary $y=4$.
On the other hand, to look for extreme points (max/min points) of $f$ on $y=4$, you can plug $4$ in for $y$ into the equation for $f$ and decide which values of $x$ between $-2 \leq x \leq 2$ make $f$ the largest and smallest. 
Once you have made a list of all possible candidates for max and min (the critical points on the interior of the region, and those found on the boundary), you then decide what the absolute max and min are for $f$ by simply checking the value of $f$ evaluated at each of those points. 
A: Personally, I should not worry about the very restrictive constraints which are given in your problem and I should just write the general problem as classically  
F = (y^2 -2 x y + x^3 - x) + a (y - x^2) + b (y - 4)    
where a and b are the Lagrange multipliers.
Writing the derivatives of F with respect to x, y, a and b leads to four equations to be solved for x, y, a and b. The only possible solutions are
{x=-2 , y=4 , a=-3/4 , b=-45/4} and {x=2 , y=4 , a=3/4 , b=-19/4}.
Now, the two extrema correspond to 26 for the first solution and 6 for the second solution.
