1. The problem statement, all variables and given/known data find the absolute extrema of f(x,y) = 2x - 2xy + y^2 in the region in the xy plane bounded by the graphs of y= x^2 and y = 1 3. The attempt at a solution first we find the first partials fx(x,y) = 2 - 2y fy(x,y) = 2y-2x 2-2y = 0 when y = 1 2y - 2x = 0 when y=x in this case y=1 so x =1 f(1,1) = 1 now we check along y = 1 f(x,1) = 2x - 2x + 1 = 1 so z = 1 along the entire line y =1 . this includes the points where y=1 intersects with y=x^2 now we check along y = = x^2 f(x, x^2) = 2x -2x^3 + x^4 f'(x,x^2) = 2 - 6x^2 + 4x^3 we need this to equal 0 2 - 6x^2 + 4x^3 = 0 1 - 3x^2 + 2x^3 = 0 now I am a little lost. I need to solve this for x but there doesn't seem to be an easy way to do this. After I solve for x I plug the corresponding point into the original function to find the minimum. (the minimum has to be along this curve because it was nowhere else and it is a bounded region) I could be doing this wrong though as these types of problems are very confusing to me .