Amy W.

asked • 03/30/17# Lagrange Multipliers

Lagrange multipliers work great for finding the minimum value of y-ax, restricted to the curve y^2-x^2=1 for any a with 0<=a<1 (what is it?) but fails quite miserably if a >=1. Use the geometry of the situation to explain this.

I already found y=+-sqrt(1/(1-a^2)) and x=+-a(sort(1/(1-a^2))), so obviously if a >=1, these are both undefined. But I don't really understand what this means geometrically.

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