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If we are to use the method of Lagrange multiplier/s in finding the absolute extrema of the function f («, у, 2) — а* + y + 2 subject to the constraints x — 2y + 32 — 4 — 0 and 2 + 3у — 4х + 5 — 0, which of the following CANNOT be the auxiliary function F(x, y, z, A, µ)? F(«, у, 2, А, м) — а4 + y + 2 +21(х — 2у + 3z — 4) — Зи(2а + Зу — 42 + 5) - | F(x, y, z, A, µ) = x* + y* + z* – X(x – 2y + 3z – 4) – µ(2x + 3y – 4z+5) - F(z, у, z, A, и) %— 2 + + + Ay(a — 2y + 32 — 4) + ря(2а + 3у — 42 + 5) F(x, y, z, A, µ) = xª + y4 + zª – (x – 2y + 3z – 4) + 4(2x + 3y – 4z + 5) - -

If we are to use the method of Lagrange multiplier/s in finding the absolute extrema of the function f («, у, 2) — а* + y + 2 subject to the constraints x — 2y + 32 — 4 — 0 and 2 + 3у — 4х + 5 — 0, which of the following CANNOT be the auxiliary function F(x, y, z, A, µ)? F(«, у, 2, А, м) — а4 + y + 2 +21(х — 2у + 3z — 4) — Зи(2а + Зу — 42 + 5) - | F(x, y, z, A, µ) = x* + y* + z* – X(x – 2y + 3z – 4) – µ(2x + 3y – 4z+5) - F(z, у, z, A, и) %— 2 + + + Ay(a — 2y + 32 — 4) + ря(2а + 3у — 42 + 5) F(x, y, z, A, µ) = xª + y4 + zª – (x – 2y + 3z – 4) + 4(2x + 3y – 4z + 5) - -

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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If we are to use the method of Lagrange multiplier/s in finding the absolute extrema of the function f(x, y, z) = x + y + z* subject to the constraints x which of the following CANNOT be the auxiliary function F(x, y, z, A, µ)? 2у + 3z — 4 O and 2x + 3y – 4z + 5 = 0, F(1, у, z, A, м) — а4 + у + 2+ 24(х — 2у + 32 — 4) — Зг(2г+ 3у — 42 + 5) - L. F(x, y, z, A, µ) = x² + y* + z* – (x – 2y + 3z – 4) – µ(2x + 3y – 4z + 5) - - F(«, у, 2, A, н) — 2 + у + +y( — 2y + 32 — 4) + рa(2x + Зу — 42 + 5) F(x, y, z, A, µ) = x* + y* + zA (x – 2y + 3z - 4) + 4 (2x + 3y – 4z + 5) –
Transcribed Image Text:If we are to use the method of Lagrange multiplier/s in finding the absolute extrema of the function f(x, y, z) = x + y + z* subject to the constraints x which of the following CANNOT be the auxiliary function F(x, y, z, A, µ)? 2у + 3z — 4 O and 2x + 3y – 4z + 5 = 0, F(1, у, z, A, м) — а4 + у + 2+ 24(х — 2у + 32 — 4) — Зг(2г+ 3у — 42 + 5) - L. F(x, y, z, A, µ) = x² + y* + z* – (x – 2y + 3z – 4) – µ(2x + 3y – 4z + 5) - - F(«, у, 2, A, н) — 2 + у + +y( — 2y + 32 — 4) + рa(2x + Зу — 42 + 5) F(x, y, z, A, µ) = x* + y* + zA (x – 2y + 3z - 4) + 4 (2x + 3y – 4z + 5) –
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