Question: User: . Lusanda has a utility function $U = 45 \text{House}^{0.4} \text{Car}^{0.5}$. The prices of the two goods, House and Car, are 2 and 3 (measured in millions of rands) for each, respectively, and Lusanda has an income of 450 million.
i. Derive the budget constrain equation
ii. State the economic condition for utility maximisation.
iii. Derive the slope of the indifference and budget constraint curves.
iv. Using the economic condition in (b) and the slopes in (c) above, find the Houses and Cars Lusanda needs to maximise his utility.
v. Depict your answer in (d) on the diagram showing the number of utils, number of Houses and Cars and intercepts of the budget constrain curve.
9. If a consumer’s utility function is given as $U = 45 X^{0.4} Y^{0.5}$, where $X$ and $Y$ are two goods worth prices 2 and 3 respectively. Given an individual’s income of 450, with the use of (Economics and Substitution method) find the quantities of $X$ and $Y$ that maximise utility ($U$).
1. **Problem Statement:**
Lusanda's utility function is given by $$U = 45 \text{House}^{0.4} \text{Car}^{0.5}$$ with prices $P_{House} = 2$ and $P_{Car} = 3$, and income $I = 450$ million.
2. **Budget Constraint Equation:**
The budget constraint states that total spending on Houses and Cars cannot exceed income:
$$2 \times \text{House} + 3 \times \text{Car} = 450$$
3. **Economic Condition for Utility Maximisation:**
Utility maximisation occurs where the marginal rate of substitution (MRS) equals the price ratio:
$$MRS = \frac{MU_{House}}{MU_{Car}} = \frac{P_{House}}{P_{Car}}$$
4. **Marginal Utilities:**
Calculate marginal utilities by differentiating $U$:
$$MU_{House} = \frac{\partial U}{\partial House} = 45 \times 0.4 \times House^{0.4 - 1} \times Car^{0.5} = 18 \times House^{-0.6} \times Car^{0.5}$$
$$MU_{Car} = \frac{\partial U}{\partial Car} = 45 \times 0.5 \times House^{0.4} \times Car^{0.5 - 1} = 22.5 \times House^{0.4} \times Car^{-0.5}$$
5. **Set MRS equal to price ratio:**
$$\frac{MU_{House}}{MU_{Car}} = \frac{18 House^{-0.6} Car^{0.5}}{22.5 House^{0.4} Car^{-0.5}} = \frac{18}{22.5} House^{-1} Car^{1} = \frac{2}{3}$$
Simplify:
$$\frac{18}{22.5} = 0.8$$
So:
$$0.8 \times \frac{Car}{House} = \frac{2}{3}$$
6. **Solve for $Car$ in terms of $House$:**
$$\frac{Car}{House} = \frac{2}{3} \div 0.8 = \frac{2}{3} \times \frac{1}{0.8} = \frac{2}{3} \times 1.25 = \frac{5}{6}$$
Thus:
$$Car = \frac{5}{6} House$$
7. **Slope of Indifference Curve:**
The slope of the indifference curve is the negative of MRS:
$$Slope_{indifference} = - MRS = - \frac{MU_{House}}{MU_{Car}} = - \frac{2}{3}$$
8. **Slope of Budget Constraint:**
From budget constraint:
$$2 House + 3 Car = 450$$
Rearranged:
$$Car = \frac{450 - 2 House}{3}$$
Slope:
$$Slope_{budget} = - \frac{2}{3}$$
9. **Find quantities of House and Car:**
Substitute $Car = \frac{5}{6} House$ into budget constraint:
$$2 House + 3 \times \frac{5}{6} House = 450$$
$$2 House + \frac{15}{6} House = 450$$
$$2 House + 2.5 House = 450$$
$$4.5 House = 450$$
$$House = \frac{450}{4.5} = 100$$
Then:
$$Car = \frac{5}{6} \times 100 = \frac{500}{6} = 83.33$$
10. **Summary:**
- Budget constraint: $$2 House + 3 Car = 450$$
- Economic condition: $$\frac{MU_{House}}{MU_{Car}} = \frac{P_{House}}{P_{Car}}$$
- Slopes: $$Slope_{indifference} = Slope_{budget} = - \frac{2}{3}$$
- Optimal quantities: $$House = 100, \quad Car = 83.33$$
11. **Diagram Description:**
- Budget line intercepts:
- When $House=0$, $Car = \frac{450}{3} = 150$
- When $Car=0$, $House = \frac{450}{2} = 225$
- Optimal point at $(House, Car) = (100, 83.33)$ lies on budget line.
- Utility at optimum:
$$U = 45 \times 100^{0.4} \times 83.33^{0.5}$$
Calculate:
$$100^{0.4} = e^{0.4 \ln 100} = e^{0.4 \times 4.605} \approx e^{1.842} \approx 6.31$$
$$83.33^{0.5} = \sqrt{83.33} \approx 9.13$$
$$U \approx 45 \times 6.31 \times 9.13 = 45 \times 57.6 = 2592$$