need help finding utility function using differentiation...?

Question

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?

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.. thanks for pointing it out

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 ...

Answer 1

No utility at the optimum should never be zero since it should be an element of the positive real numbers.

I think what you are saying is that the derivative of utility is 0 at the optimum.

In your example, however, the optimum could only be at zero because of your functional form. The only place du/dy could be zero is at zero. Your functional form is convex, the only way to optimize to a maximum given your utility function would be to have concave resources. This would be a little strange. It would imply that you could, for example, purchase 1000 units or 10,000 units but not necessarily any number in between.

Further, you did not provide a stated constraint, such as a budget constraint. Against what would you optimize? Infinite resources? Then you should use an infinite amount.

Consider a Cobb-Douglass utility function if you are looking for an example. It is a versatile form and easy to understand. Or consider quasi-linear utility. Again, it is versatile and is an easy one to explain. Set the fixed consumption as toilet paper, since the demand for toilet paper is price invariant. It only is determined by biological factors. So in practice, toilet paper becomes the numeraire of the economy.

Just a note, there is an error in your calculus.

If U(x,y)=2x^3*y then you are dealing with partial derivatives. Rather than use the du/dx form, I will use pu/px for the partial form since the del operator is unavailable.

pU/px=6yx^2
pU/py=2x^3

dy/dx=(-1)pu/px/(pu/py)

So
dy/dx=(-1)*(6yx^2)/(2x^3)

so
dy/dx=-3y/x

dy/dx is your marginal rate of substitution.

Finally marginal utility is the pu/px and pu/py.

If marginal utility is 2x^3*y then you have a different problem present.

Answer 2

if u(x,y) = 2x^3y , yes du/dy and du/dx you calcuated correctly

but i dont know how you computed dx/dy,
i guess you did
dx/dy = du/dy / du/dx = x/3y, but that is incorrect.

Answer 3

u r right