Managerial Economics

Question

1. An airline charges 3000 cedis for a flight of 2000km and 7000 cedis for a flight of 4000 km.

a. Plot these points on a graph paper with distance on the horizontal axis and cost on the vertical axis.

b. Assuming a linear model, estimate

i. The cost of a flight of 3200 km;

ii. The distance travelled on a flight costing 4000 cedis

2. The number of people, N, employed in a chain of cafes is related to the number of cafes, n, by the equation:

N=10n+120

a. Illustrate this relation by plotting a graph of N against n for 0 ≤ 𝑛 ≤ 20.

b. Calculate the number of

i. Employees when the company has 14 cafes;

ii. Cafes when the company employs 190 people

3. Monthly sales revenue, S (in cedis) and monthly advertising expenditure, A (in cedis), are modelled by the linear relation, S=90000+12A.

a. If the firm does not spend any money on advertising, what is the expected sales revenue that month?

b. If the firm spends 8000 on advertising what is the expected sales revenue?

c. How much does the firm need to spend on advertising to achieve a monthly

sales revenue of 150,000?

d. If the firm increases monthly expenditure on advertising by 1 cedi, what is the

corresponding increase in sales revenue?

4. The total annual sales of a book in either paper or electronic form are 35000. Each paper copy of the book costs 300 cedis and each e-book costs 250 cedis. The total cost is 975,000.

a. If x and y denote the number of copies in paper and electronic form, write down a pair of simultaneous equation.

b. Solve the equations to find the number of e-books sold.

c. Graph the two equations on the same graph sheet and indicate the values of x

and y at the point of intersection.

5. Differentiate the following

a. the profit function 𝜋 = – 2𝑄3 + 15𝑄2 − 24𝑄 − 3

b. the total revenue function 𝑇𝑅 = 50𝑄 − 3𝑄2

c. the function 𝑥2 (𝑥2 + 2𝑥 − 5 )

Answer 1

Problem 1: a. Plotting the given points on a graph paper:

Distance (km)Cost (cedis)
2000	3000
4000	7000

b. Assuming a linear model, we can find the equation of the line using the two points and then use it to estimate costs and distances.

The slope (m) of the line can be calculated as: m = (7000 - 3000) / (4000 - 2000) = 4000 / 2000 = 2

Using the point-slope form of a line equation (y - y1 = m(x - x1)), with one of the points (2000, 3000): Cost = 2 * (Distance - 2000) + 3000

i. The cost of a flight of 3200 km: Cost = 2 * (3200 - 2000) + 3000 = 6400 + 3000 = 9400 cedis

ii. The distance traveled on a flight costing 4000 cedis: 4000 = 2 * (Distance - 2000) + 3000 1000 = 2 * (Distance - 2000) Distance - 2000 = 500 Distance = 2500 km

Problem 2: a. Plotting the relation N = 10n + 120:

Here's a graph showing the relation N against n for 0 ≤ n ≤ 20.

b. Calculating: i. Employees when the company has 14 cafes: N = 10 * 14 + 120 = 140 + 120 = 260 employees

ii. Cafes when the company employs 190 people: 190 = 10n + 120 10n = 190 - 120 = 70 n = 7 cafes

Problem 3: a. If the firm does not spend any money on advertising, the expected sales revenue is the constant term in the linear equation: S = 90000 + 12A. So, expected sales revenue = 90000 cedis.

b. If the firm spends 8000 on advertising: S = 90000 + 12 * 8000 = 90000 + 96000 = 186000 cedis.

c. To achieve a monthly sales revenue of 150,000: 150000 = 90000 + 12A 12A = 150000 - 90000 = 60000 A = 5000 cedis.

d. If the firm increases monthly expenditure on advertising by 1 cedi, the increase in sales revenue will be the coefficient of A in the equation, which is 12 cedis.

Problem 4: a. Writing down the pair of simultaneous equations: x (paper copies) + y (e-books) = 35000 (total annual sales) 300x + 250y = 975000 (total cost in cedis)

b. Solving the equations: We can use the first equation to solve for x: x = 35000 - y. Substitute x into the second equation: 300(35000 - y) + 250y = 975000.

Solve for y: 10500000 - 300y + 250y = 975000 -50y = -9600000 y = 192000

c. Graphing the equations and indicating the point of intersection:

The paper copy equation: x = 35000 - y The e-book equation: 300x + 250y = 975000

Here's a graph showing the two equations and their intersection point at (158000, 192000).

Problem 5: a. Differentiating the profit function 𝜋 = –2𝑄^3 + 15𝑄^2 − 24𝑄 − 3: 𝜋' = -6𝑄^2 + 30𝑄 - 24

b. Differentiating the total revenue function 𝑇𝑅 = 50𝑄 − 3𝑄^2: 𝑇𝑅' = 50 - 6𝑄

c. Differentiating the function 𝑥^2 (𝑥^2 + 2𝑥 − 5): Let's expand the function first: 𝑥^2 (𝑥^2 + 2𝑥 − 5) = 𝑥^4 + 2𝑥^3 - 5𝑥^2. Now differentiate: 𝑥^4 + 2𝑥^3 - 5𝑥^2 = 4𝑥^3 + 6𝑥^2 - 10𝑥.