Question

The Implicit Function Theorem and the Marginal Rate of Substitution (4 Points) 3 An important result from multivariable calculus is the implicit function theorem which states that given a function f (x,y), the derivative of y with respect to a is given by where of/bx denotes the partial derivative of f with respect to a and af/ay denotes the partial derivative of f with respect to y. Simply stated, a partial derivative of a multivariable function is the derivative of that function with respect to one particular variable, treating all other variables as constant. For example, suppose f(x, y)-zy2 . To compute the partial derivative of f with respect to z, we treat y as a constant, in which case we obtain af/ox-y2 and to compute the partial derivative of f with respect to y, we treat z as a constant, in which case we obtain df/dy-2xy. We have described the slope of an indifference curve as the marginal rate of substitution between the two goods. Imagining that c2 is plotted on the vertical axis and ci plotted on the horizontal axis, compute the marginal rate of substitution for the following utility functions a. (1 Point) u(a, c2)-m(c1) + ln(c2) b. (1 Point) u(c1,c2)-c +c2 c. (2 Points) u(C1, C2) c c1-α, where α Ε (0,1) is some constant.

Answer 1

a. u(C_1, C_2) = Ln(C_1) + (C_1) + Ln(C_2) Now partial differential u/ partial differential C_1 = 1/C_1 and partial differential u/ partial differential C_2 = 1/C_2 Therefore partial differential C_2/ partial differential C_1 = partial differential u/ partial differential C_1/ partial differential u/ partial differential C_2 partial differential C_2/ partial differential C_1 = C_2/ C_1 b u(C_1, C_2) = squareroot C_1 + squareroot C_2 Now partial differential u/ partial differential C_1 = 1/2 squareroot C_1 and partial differential u/ partial differential C_2 = 1/ 2 squareroot C_2 Therefore partial differential C_2/ partial differential C_1 = partial differential u/ partial differential C_1/ partial differential u/ partial differential C_2 partial differential C_2/ partial differential C_2 = squareroot C_2/C_1 c. u(C_1, C_2) = C_1^a C^1 - a_2 Now, partial differential u/ partial differential C_1 = a C1^a-1 C2^1-a and partial differential u/ partial differential C_2 = (1 - a)C_1^a C^-a_2 Therefore partial differential C^2/ partial differential C_1 =