Question

7. An individual's preferences are represented by the utility function Ua, y) 4xy x. Which of the following statements is false? a. The marginal utility of x increases as x increases, holding y constant. b. Preferences are monotonic in both goods. c. The indifference curves slope downward at a decreasing rate. d. The marginal rate of substitution ofx for y increases as y increases, holding x constarnt e. The consumer is willing to give up decreasing amounts of good y for each additional unit of good x in order to remain as well off.

Answer 1

7. The correct option is

• a. The marginal utility of x increases as x increases, holding y constant.

We have the MUs as $\dpi{80} MU_x = \frac{\partial U}{\partial x}$ or $\dpi{80} MU_x = 4y+1$ and $\dpi{80} MU_y = \frac{\partial U}{\partial y}$ or $\dpi{80} MU_y = 4x$ . As $\dpi{80} MU_x = 4y+1$ , holding y constant, MU of x will also be a constant, and would not increases if x increase. Only the first option is hence false.

Preferences are monotonic for x as for bundle (x1,y1) and (x2,y1), where x2>x1, we have $\dpi{80} U_1 = 4x_1y_1 + x_1$ and $\dpi{80} U_2 = 4x_2y_1 + x_2$ , and hence $\dpi{80} U_2 - U_1 = 4x_2y_1 + x_2 - (4x_1y_1 + x_1)$ or $\dpi{80} U_2 - U_1 = 4y_1(x_2 - x_1) + (x_2 - x_1)$ or $\dpi{80} U_2 - U_1 = (x_2 - x_1)(4y_1 + 1) > 0$ (since x2>x1) or $\dpi{80} U_2 > U_1$ . Similarly, preferences are monotonic for y as for bundle (x1,y1) and (x1,y2), where y2>y1, we have $\dpi{80} U_1 = 4x_1y_1 + x_1$ and $\dpi{80} U_2 = 4x_1y_2 + x_1$ , and hence $\dpi{80} U_2 - U_1 = 4x_1y_2 + x_1 - (4x_1y_1 + x_1)$ or $\dpi{80} U_2 - U_1 = 4x_1y_2 - 4x_1y_1$ or $\dpi{80} U_2 - U_1 = 4x_1(y_2 - y_1) > 0$ or $\dpi{80} U_2 > U_1$ .

The slope of the indifference curve is the MRS itself. The MRS would be as $\dpi{80} MRS = -\frac{MU_x}{MU_y}$ , and we have $\dpi{80} MRS = -\frac{4y+1}{4x}$ . As the MRS is negative, the IC is downward sloped indeed. Also, for $\dpi{80} \frac{\mathrm{d} y}{\mathrm{d} x} = -\frac{4y+1}{4x}$ , we have $\dpi{80} \frac{\mathrm{d}^2 y}{\mathrm{d} x^2} = -\frac{\mathrm{d} }{\mathrm{d} x}(\frac{4y+1}{4x})$ or $\dpi{80} \frac{\mathrm{d}^2 y}{\mathrm{d} x^2} = - (- \frac{4y+1}{4x^2})$ or $\dpi{80} \frac{\mathrm{d}^2 y}{\mathrm{d} x^2} = \frac{4y+1}{4x^2}$ . If x increase and y decrease, the rate of change of MRS $\dpi{80} \frac{\mathrm{d}^2 y}{\mathrm{d} x^2} = \frac{4y+1}{4x^2}$ would decrease. Hence, the slope decreases at a decreasing rate.

Holding x constant in $\dpi{80} MRS = -\frac{4y+1}{4x}$ , if y is increased, so would the |MRS| (absolute value of MRS).

As the MRS slopes downward, that means to get a marginal unit of x, y should be decreased by the amount of MRS to remain as well off as before.