Answer to Question #167471 in Microeconomics for SYEDA ZAINAB

Question

Q: A consumer's utility function is given by the expression: U = (0.6X^0.5+ 0.4Y^0.5)².

• Determine the marginal utility functions for each commodity. Does marginal utility decrease when consumption increases?
• Assuming that the price of good X is Rs 15 and the price of Y is Rs 6, write the equation of the budget line and plot it when income is Rs 450. What is its slope? What does it indicate?
• Calculate the marginal rate of substitution of Y for X and interpret its economic meaning. Write the equation showing the consumer's equilibrium condition.
• Obtain the equilibrium values of X and Y.
• Find the expressions for change in MU_x due to an increase in Y and change in MU_y due to an increase in X.

Answer 1

u=(0.6X^0.5+0.4Y^0.5)²

a. MU_x=2(0.6X^0.5+0.4Y^0.5)(0.3X^-0.5)

MU_y=2(0.6X^0.5+0.4Y^0.5)(0.2Y^-0.5)

Marginal utility decreases as consumption of the commodity increases.

b. if price is of X is RS15 ; price of Y is RS6 ; and income is RS450

then the equation of the of the consumer is given as:

$450=15X+6Y$

and the equation of the budget line is

$Y=75-2.5X$ 

and the slope = -2.5

The slope shows the ratio of the prices

c. MRS_{y for x} = $\frac{2(0.6X^{0.5}+0.4Y^{0.5})(0.3X^{-0.5})}{2(0.6X^{0.5}+0.4Y^{0.5})(0.2Y^{-0.5})}$

= $\frac{0.3X^{-0.5}}{0.2Y^{-0.5}} ​$

= $\frac{3}{2}{(\frac{Y}{X})}^{0.5}$

The consumer's equilibrium condition is given as

$\frac{MUx}{MUy}=\frac{Px}{Py}$

$\frac{3}{2}{(\frac{Y}{X})}^{0.5} = \frac{15}{6}$

d. $\frac{3}{2}{(\frac{Y}{X})}^{0.5} = \frac{15}{6}$

$(\frac{Y}{X})^{0.5}=\frac{15}{6}×\frac{2}{3}$

$(\frac{Y}{X})^{0.5}=\frac{5}{3}$

$\frac{Y}{X}=(\frac{5}{3})^{2}$

Y = 2.78X

Substituting the value of Y in the budget line:

$450=15X+6Y$

$450=15X+6(2.78X)$

450=15X + 16.7X

450=31.7X

X= 14.2

Y= 2.78(14.2)

Y= 39.8

The equilibrium values for X and Y are (14.2,39.8)

e. i) $\Delta{MUx} =\frac{\delta{MUx}}{\delta{y}}{\Delta{Y}}$

= $0.12({XY})^{-0.5}\Delta{Y}$

ii) $∆MUy=\frac{\delta{MUy}}{\delta{x}}\Delta{X}$

= 0.12$(XY)^{-0.5}\Delta{X}$