Lagrange Multipliers (and finding extrema of a function with two restraints)

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SUMMARY

The discussion focuses on finding the extrema of the function f(x,y,z) = x + y + z under the constraints x² - y² = 1 and 2x + z = 1 using Lagrange multipliers. The gradient of f is established as (1, 1, 1), which equals λ₁(2x, -2y, 0) + λ₂(2, 0, 1). The analysis reveals that λ₁ = -1/(2x) = -1/(2y) implies x = y, contradicting the constraint x² - y² = 1. This indicates that absolute maxima or minima may not exist within the defined constraints, suggesting the extrema could lie on the boundary of the feasible region.

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  • Basic calculus concepts related to extrema
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Black Orpheus
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I need to find the extrema of f(x,y,z)=x+y+z subject to the restraints of x^2 - y^2 = 1 and 2x+z = 1. So the gradient of f equals (1,1,1) =
lambda1(2x,-2y,0) + lambda2(2,0,1). Solving for the lambdas I found that lambda1 = -1/(2x) = -1/(2y), or x=y. But this isn't possible if x^2 - y^2 = 1. Does this mean that there are no absolute max or min, or am I doing something wrong?
 
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It may mean that the max or min is on the boundary of the set.
 
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