Mathematics
Mathematics, 16.08.2020 01:01, walterzea70

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 Ξ» = 0. Now, the constraint g(x, y) = 2x2 + 4y2 = 12 gives us y = Β± $$√3 . If instead, we begin with Ξ» = y, then x2 = 4Ξ»y implies that x2 = 4y2. Now 12 = g(x, y) = 2x2 + 4y2 = 12y2, and so y = Β± $$1 . Step 5 If y = 1, then 2x2 + 4y2 = 12 tells us that x = Β± $$2 . The same is true if y = βˆ’1. Step 6 We now know that f has possible extreme values at the six points A(0, 3 ), B(0, βˆ’ 3 ), C(2, 1), D(βˆ’2, 1), E(2, βˆ’1) and F(βˆ’2, βˆ’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)

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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...

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