can someone help me with differentiation?
Utility (at optimum) is always = 0
if for eg, the utility function = 2x³y
du/dy = 2x³
and du/dx = (3)(2)x²y = 6x²y
and dx/dy = 3y/x
am i correct with my example?
2007-08-11
04:29:30
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3 answers
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asked by
Anonymous
in
Social Science
➔ Economics
sorry, i forgot to mention, du/dx = MUx , and du/dy = MUy, and so MUx/MUy is what i was really trying to say, not dx/dy , yup..
and if the power is too small, im writing it again.
the utility function = 2x^3y
du/dy = 2x^3
and du/dx = (3)(2)x^2y = 6x^2y
and dx/dy = 3y/x
and its' not "utility = 0", but "marginal utility" = 0 ...
2007-08-11
05:24:58 ·
update #1