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 Solve the following constrained optimization problems. These problems, while abstract, illustrate how some care must be used when applying Lagrange's method. Sketching pictures should help. Although only the maximum is asked for in each problem, make sure that you also find the arguments that maximise the objective functions. max 152 + 112 (u,v)E]R2 subject to \"2 + 1m + v2 = 3 2. max 21; + 31) [u,v)E]Ri subject to +=5