If x = 0 and y = 0, then g(x, y) = 0, which contradicts g(x, y) = 12. So if x = 0, then we must have λ = 0. Now, the constraint g(x, y) = 2x2 + 4y2 = 12 gives us y = ± $$√3 . If instead, we begin with λ = y, then x2 = 4λy implies that x2 = 4y2. Now 12 = g(x, y) = 2x2 + 4y2 = 12y2, and so y = ± $$1 . Step 5 If y = 1, then 2x2 + 4y2 = 12 tells us that x = ± $$2 . The same is true if y = −1. Step 6 We now know that f has possible extreme values at the six points A(0, 3 ), B(0, − 3 ), C(2, 1), D(−2, 1), E(2, −1) and F(−2, −1). After evaluating f(x, y) at these points, we find the maximum occurs at points C and B , and the minimum at points E and C . Submit Skip (you cannot come back)