Part 1: Locate all the relative extrema and saddle points of f(x, y) = 3x - 2xy + y+8y. First of all, find fx and fy, then solve fx = 0 and fy = 0 for critical points. fx(x, y) = X 6x+y^2-6y == X 3x^2+y^2-2x+8y fy(x, y) = Solving fx = 0 and fy = 0 you get only one critical point (0,0) where (input functions in increasing alphabetic order, put constant term before variables/functions) X-8 . Yo . (it is not hard, all you need is basic elimination of one unknown then solve for the remaining unknown) To decide if (0, 0) is min/max/saddle, you apply 2nd order derivative test: 6 x 2y fax (20,90) fyy (xo, yo) fxy (xo, yo): min Therefore, (0, 0) is a local (input one of min, max, saddle, none) Part 2: The above is an unconstrained optimization problem. If there is a constraint, say g(x, y) = x + y = 0. Then clearly the critical point (0,0) found solely from fx = 0 and fy = 0 cannot satisfy the given constraint. So it cannot be the critical point of the constraint optimization problem. To take into account of the constraint, we need to introduce the Lagrange multiplier. Let L(x,y,A) = f(x, y) - Ag(x, y) = 3x - 2xy + y + 8y = X(x + y). == Then, we find critical point(s) from L (x, y, A) = Ly(x, y, A) = Lx(x, y, ) = 0, which naturally takes care of the constraint. X (no answer) = L(x, y, A) 0 leads to = X (no answer) = Ly(x, y, A) 0 leads to = (input in increasing alphabetic order of variables; input the constant term before any variable terms) (no answer) Lx(x, y, A) = 0 leads to y = (input a function of x) Substitute the equation you get from Lx (x, y, ) = 0 into the two equations you get from L (x, y, A) = Ly(x, y, A) = 0, you can solve for the new critical point, its x- X (no answer) coordinate is (input a most simplified fraction). (no answer) The new critical point clearly satisfies the constraint since its y-coordinate is (input a most simplified fraction).