Absolute extrema of the function $f(x,y)=2xy-x-y$ Find the absolute extrema of the function $$f(x,y)=2xy-x-y$$ over the region of the $xy$-plane bounded by the parabola $y=x^2$ and the line $y=4.$ 
I was wondering if I needed to use Lagrange multipliers to solve this problem and if I do, how would I go about solving this problem? If someone could help me, that would be great! Thanks
 A: To return to this problem, the reason the Lagrange-multiplier method won't be enough (or possibly even necessary) to solve the problem is that it "looks" for points on a level curve or surface of the function to be extremized where the normal vector is parallel (or anti-parallel) to a normal vector to the constraint curve or surface.  So it can help with the boundary of a region, but may not tell us much about the interior.
When a closed region is specified on which to locate extrema, to be thorough, we follow a procedure analogous to what we use in single-variable calculus for locating extrema on a closed interval:  find critical points in the interior of the region using derivatives; find critical points on bounding curves of the region (often also using derivatives or Lagrange multipliers); evaluate the function at vertices; compare the values of the function among these results.
The first partial derivatives of the function $ \ f(x, \ y) \ = \ 2xy \ - \ x \ - y \ $ are  $ \ f_x \ =  2y \ - \ 1 \ $ and $ \ f_x \ =  2x \ - \ 1 \ $ ; these are both equal to zero at $ \ ( \ \frac{1}{2} , \ \frac{1}{2} \ ) \ $ , so this is the only critical point within the parabolic region.  The value of the function there is $ \ f( \ \frac{1}{2} , \ \frac{1}{2} \ ) \ = \ \frac{1}{2} \ - \ \frac{1}{2} \ - \ \frac{1}{2} \ = \ -\frac{1}{2} \ $ .
The two "bounding curves" of the region are the parabola $ \ y \ = \ x^2 \ $ and the "horizontal" line $ \ y \ = \ 4 \ $ .  We can use these equations to reduce the extremization problem to one with single-variable functions:
$ \mathbf{on \ y = 4 :} \quad f(x, \ 4) \ = \ 8x \ - \ x \ - 4 \ = \ 7x \ - \ 4 \ \ \Rightarrow \ \ \frac{df}{dx} \ = \ 7 \ \ ; $
$ \mathbf{on \ y = x^2 :} \quad   f(x, \ x^2) \ = \ 2x^3 \ - \ x \ - \ x^2  \ \ \Rightarrow \ \ \frac{df}{dx} \ = \ 6x^2 \ - \ 2x \ - \ 1 \ \ . $
There is no critical point on the line, since the derivative is never zero there; the fact that the function  $ \ f(x, \ 4) \ $ increases linearly with $ \ x \ $ suggests that the endpoints of this boundary segment may be extrema.  On the parabola, $ \frac{df}{dx} \ = \ 6x^2 \ - \ 2x \ - \ 1 \ = \ 0 \ $ for  $ \ x \ = \ \frac{2 \ \pm \ \sqrt{28}}{12} \ = \ \frac{1 \ \pm \ \sqrt{7}}{6} \   $ . The parabola intersects the line at $ \ ( \pm 2, \ 4) \ $ , so both of these values of $ \ x \ $ do correspond to locations on the restricted parabolic boundary:  $ \  \left(  \frac{1 \ - \ \sqrt{7}}{6}  , \  \frac{4 \ - \ \sqrt{7}}{18} \right) $ and $ \  \left(  \frac{1 \ + \ \sqrt{7}}{6}  , \  \frac{4 \ + \ \sqrt{7}}{18} \right)  \ $ .  At these points, the value of our function is
$$ f \left(  \frac{1 \ - \ \sqrt{7}}{6}  , \  \frac{4 \ - \ \sqrt{7}}{18} \right)  \ = \ \frac{7 \sqrt{7} \ - \ 10}{54} \ \approx \ 0.158 $$
and
$$ f \left(  \frac{1 \ + \ \sqrt{7}}{6}  , \  \frac{4 \ + \ \sqrt{7}}{18} \right)  \ = \ -\frac{7 \sqrt{7} \ + \ 10}{54} \ \approx \ -0.528 \ \ . $$
Finally, we need to evaluate the function at the vertices of the parabolic region:
$$ f (  -2  , \  4 )  \ = \ -16 \ - \ (-2) \ - 4 \ = \ -18 \ \ \  \text{and} \ \ \ f (  2 , \  4 )  \ = \ 16 \ - \ 2 \ - 4 \ = \ 10 \ \ , $$
which are just the values that $ \ 7x \ - \ 4 \ $ gives us.  These are by far the greatest and least values of the function, so they are the absolute extrema for the region.
Had we applied Lagrange multipliers, we would need to treat the two portions of the boundary separately.  The gradient of our function is $ \ \nabla f \ = \ \langle \ 2y \ - \ 1 \ , \ 2x \ - \ 1 \ \rangle \ $ .  The constraint function for the parabolic curve is $ \ g(x,y) \ = \ y \ - \ x^2 \ $ , so its gradient is $ \ \nabla g \ = \ \langle \ -2x \ , \ 1 \ \rangle \ $ .  From the Lagrange equation $ \ \nabla f \ = \ \lambda \nabla g \ $ , we obtain
$$ 2y \ - \ 1 \ = \ \lambda \ \cdot \ ( \ -2x \ ) \ \ , \ \ 2x \ - \ 1 \ = \ \lambda \ \cdot \ 1 $$
$$ \Rightarrow \ \ \lambda \ = \ -\frac{2y \ - \ 1}{2x} \ = \ 2x \ - \ 1 \ \ \Rightarrow \ \ 2y \ = \ 1 \ + \ 2x \ - \ 4x^2 \ \ ; $$
applying the constraint  $ \ y \ = \ x^2 \ $ gives us the quadratic equation we found earlier, $ \ 6x^2 \ - \ 2x \ - \ 1 \ = \ 0 \ $ , leading to the extrema on the curve that we determined above.
As for the horizontal line, $ \ y \ = \ 4 \ $ , Lagrange multipliers will not give us anything meaningful in this case.  It would seem that we would use a constraint function $ \ h(x, \ y) \ = \ y \ - \ 4 \ $ , for which the gradient is $ \ \nabla h \ = \ \langle \ 0, \ 1 \ \rangle \ $ .  The consequent equations from $ \ \nabla f \ = \ \lambda \nabla g \ $ are $ 2y \ - \ 1 \ = \ \lambda \ \cdot \ 0 \ \ , \ \ 2x \ - \ 1 \ = \ \lambda \ \cdot \ 1 $ .  But since we require $ \ y \ = \ 4 \ $ , this set of equations has no solution, implying that there is no level curve of $ \ f(x, \ y) \ $ which has a normal in the "vertical direction" where it touches this line.  While this confirms that there is no critical point for the function along that line segment, it doesn't do much else for us.
A: I made a plot to get familiar with problem. 

$f$ is the hyperbolic plane in red. 
The parabola is extended in $z$-direction into the green surface. 
The line $y=4$ is extended to the blue plane.
My guess is that the maximum is front left, the minimum front right. 
The intersection of $f$ and parabolic cylinder is determined by
$$
f(x,y) = 2 x y - x - y = z \\
x^2 - y = 0
$$
Along the parabola we get this height:
$$
g(x) = f(x, x^2) = 2 x^3 - x - x^2
$$
Along the line $y=4$ we get this height:
$$
h(x) = f(x, 4) = 8 x - x - 4 = 7x - 4
$$
The extrema for $x \in [-2, 2]$ are 
$$
g(2) = 16 - 2 - 4 = 10 \quad h(2) = 10 \\
g(-2) = -16 + 2- 4 = -18 \quad h(-2) = -18
$$
