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maxr₁ +37₂ ZER² s.t. x² + ax=10

maxr₁ +37₂ ZER² s.t. x² + ax=10

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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we solved this for a = 1. We found that z*(a = 1) = (1,3) is a solution with Lagrange multiplier X* (a = 1) = ¹/2 and a maximum value f*(a = 1) = f(x*(1)) = f(1, 3) = 10. We now want to use the envelope theorem to estimate the maximum value at a = 1.01, so to estimate f*(1.01). Firs,t note that the NDCQ condition is satisfied whenever a 0 (HW: Verify this!), and the Lagrangian is given by: maxr₁ +372 ZER² s.t. x² + ax = 10 L ((x; a); λ) = f(x; a)-X (g(x; a)- b) lam1 By the envelope theorem: Əf*(a) да An estimate for f*(1.01) is therefore given by = ac да 84 = x₁ + 3x₂ −X(x² + ax-10) |z=2*(a),A°=1/2 = −22²2=2(1)-(1.3) =
Transcribed Image Text:we solved this for a = 1. We found that z*(a = 1) = (1,3) is a solution with Lagrange multiplier X* (a = 1) = ¹/2 and a maximum value f*(a = 1) = f(x*(1)) = f(1, 3) = 10. We now want to use the envelope theorem to estimate the maximum value at a = 1.01, so to estimate f*(1.01). Firs,t note that the NDCQ condition is satisfied whenever a 0 (HW: Verify this!), and the Lagrangian is given by: maxr₁ +372 ZER² s.t. x² + ax = 10 L ((x; a); λ) = f(x; a)-X (g(x; a)- b) lam1 By the envelope theorem: Əf*(a) да An estimate for f*(1.01) is therefore given by = ac да 84 = x₁ + 3x₂ −X(x² + ax-10) |z=2*(a),A°=1/2 = −22²2=2(1)-(1.3) =
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