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Engineering Calculus Notes 350

# Engineering Calculus Notes 350 - 338 CHAPTER 3 REAL-VALUED...

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Unformatted text preview: 338 CHAPTER 3. REAL-VALUED FUNCTIONS: DIFFERENTIATION We can solve the ﬁrst three equations for λ2 and eliminate it: (1 − λ1 )x = (1 − λ1 )y = 2z. The ﬁrst of these equations says that either λ1 = 1 or x = y. If λ1 = 1, then the second equality says that z = 0, so y = 1 − x. In this case the ﬁrst constraint gives us x2 + (1 − x)2 = 4 2x2 − 2x − 3 = 0 √ 1 x = (1 ± 7) 2 √ 1 y = (1 ∓ 7) 2 yielding two relative critical points, at which the function f has value f √1 √ 1 (1 ± 7), (1 ∓ 7), 0 2 2 = 9 . 4 If x = y , then the ﬁrst constraint tells us x2 + x2 = 4 √ x=y=± 2 and then the second constraint says z = 1 − 2x √ =1∓2 2 ...
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