Consider the function .7: _2 Confirm that this function has a critical point at (x y) = (1 , 1) and then determine whether it corresponds to a local minimum, local maximum or saddle point. The point (z,y) = (1,1) is a local A minimum, because the Hessian is negative definite there. A- minimum, because the Hessian is positive definite there. .A; maximum, because the Hessian is negative definite there. -A: maximum, because the Hessian is positive definite there. It can be verified that argmaxmelo'l: = {1}. Use this fact to find the solution to the maximisation problem 4x 1~/13) 1 4. 32013:] (5+ 0 7