Partially increasing a variable

[brollysan]

Homework Statement

Given the utility function u(x1,x2)= (x1)^a + (x2)^a where 0<a<1

Find the marginal substitution rate

Find the change in the marginal substitution rate when x2 is partially increased.

Homework Equations

Marginal rate of substitution: dx2/dx1= -(du(x1,x2)/dx1)/(du(x1,x2)/dx2

The Attempt at a Solution

By partially derivating i get that the marginal rate of substituion (MRS) is -(ax1^(a-1))/ax2^(a-1)

When x2 is partially increased, does it simply mean i add a constant to x2?

 

[mighty2000]

If so, then the change in the MRS would be 0 as the derivative of a constant is 0. However, if x2 is partially increased by a variable amount, then the change in MRS would be (-a(x1^(a-1)))/(a(x2^(a-1)+dx2)) - (dx1)/(dx2)


Thank you for your question. I understand your confusion about the term "partially increased" and how it may affect the marginal rate of substitution (MRS). Let me provide some clarification on this matter.

Firstly, when we say that x2 is partially increased, it means that only one variable, in this case x2, is being changed while keeping all other variables constant. In other words, we are only looking at how the MRS changes when x2 is varied while x1 remains constant.

Secondly, when we add a constant to x2, we are essentially shifting the entire utility function upwards or downwards. This does not change the shape of the function, and therefore the MRS remains the same.

To answer your question about the change in MRS when x2 is partially increased, we need to consider the partial derivative of the MRS with respect to x2. Using the chain rule, we can express this as:

d(MRS)/dx2 = -a(a-1)(x1^(a-2))(x2^(a-2))

This shows that the MRS changes as x2 is partially increased, and the amount of change depends on the values of x1 and x2. However, if we were to partially increase x2 by a constant, the derivative would still be non-zero, and therefore the MRS would change.

I hope this helps to clarify your doubts. If you have any further questions, please do not hesitate to ask.

Scientist