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) - -

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