ECON201 Final Exam 2024 Solution Keys
Solution to Questions 1, 2 and 3 (Vladimir)
Question 1 25 Marks
Let us first solve the utility maximization problem with general income and prices. The
tangency condition is MUx / MUy = px / py. Given the above utility function, this condition
takes the form y / x = px / py. This is equivalent to px*x = py*y. The budget constraint is px*x +
py*y = I. Solving the two equations simultaneously yields the following utility-maximizing
amounts for x and y: x = I / (2*px) and y = I / (2*py).
a) 6 marks. When px = 2, py = 6 and I = 60, we get x = 60/4 = 15 and y = = 5. The
level of utility is U = 5*15 = 75.
b) 6 marks. The tax rate t satisfies 10 = 30/(2 + t). Solving this equation for t yields t = 1.
c) 6 marks. The resulting tax revenue will be R = t*x = 1*10 = 10.
d) 7 marks. The lump-sum tax R will satisfy 10 = (60 – R) / 4. Solving this equation for
R yields R = 20.
Question 2 25 Marks
a) 6 marks. Cost-minimizing input bundles must satisfy the tangency condition MPL / MPK = w
/ r. Given the above production function, we have MPL / MPK = K0.5 / L0.5. Thus, the tangency
condition implies K0.5 / L0.5 = . Hence, the cost minimizing labour-capital ratio is L / K =
4.
b) 6 marks. The tangency condition gives us L = 4*K. Plug this in the production function to
get K0.5 +2*K0.5 = q. Simplification yields K0.5 = q/3. Hence, K = q and L = 4*q.
Therefore, the long-run cost is C = w*L + r*K = 1*4*q + 2*q =2*q.
c) 6 marks. If K = 1, the amount of labour that is required to produce q units of output must
satisfy 10.5 + L0.5 = q. Solving for L yields L = (q – 1)2. Therefore, the short-run cost will be
CSR = w*L + r*K = 1*(q -1)2 + 2*1 = q2 – 2*q + 3.
d) 7 marks. One way to solve this question is to realize that the short run-cost will be equal to
the long-run cost when in the short-run value of K is equal to its long-run optimum. In the
short run, K is fixed at 1. From part (b), the long-run optimal value of K is K = q. Hence,
q will satisfy 1 = q. Solving this equation yields q = 3.
Question 3 25 Marks
a) 6 marks. The marginal cost is MC = y2 – 20y + 105. The average total cost is ATC = y2
/ 3 – 10y + 105 + 5 / y. The average variable cost is AVC = y – 10y + 105.
b) 6 marks. The profit-maximizing output of this firm solves p = MC. Thus, we need to
solve 105 = y2 – 20y + 105 = 105. This equation has two roots: y = 0 and y = 20. If y =
0, the firm’s profit will be – 5. If y = 20, the firm’s profit will be 3985/3. Hence, the
profit-maximizing output level is y = 20.
c) 6 marks. The shutdown price is the minimum of the average variable cost. From part
A, we have AVC = y – 10y + 105. The value of y which minimizes AVC solves
dAVC / dy = 0. Differentiation gives us 2*y / 3 – 10 = 0. Solving this equation yields y
= 15. Substitute y = 15 in AVC to get AVC = 30. Hence, the shutdown price is 30.
1.
, d) 7 marks. The lowest positive amount of output that this firm will supply is the output
that minimizes AVC. In part (c), we found that this output is y = 15.
1.
Solution to Questions 1, 2 and 3 (Vladimir)
Question 1 25 Marks
Let us first solve the utility maximization problem with general income and prices. The
tangency condition is MUx / MUy = px / py. Given the above utility function, this condition
takes the form y / x = px / py. This is equivalent to px*x = py*y. The budget constraint is px*x +
py*y = I. Solving the two equations simultaneously yields the following utility-maximizing
amounts for x and y: x = I / (2*px) and y = I / (2*py).
a) 6 marks. When px = 2, py = 6 and I = 60, we get x = 60/4 = 15 and y = = 5. The
level of utility is U = 5*15 = 75.
b) 6 marks. The tax rate t satisfies 10 = 30/(2 + t). Solving this equation for t yields t = 1.
c) 6 marks. The resulting tax revenue will be R = t*x = 1*10 = 10.
d) 7 marks. The lump-sum tax R will satisfy 10 = (60 – R) / 4. Solving this equation for
R yields R = 20.
Question 2 25 Marks
a) 6 marks. Cost-minimizing input bundles must satisfy the tangency condition MPL / MPK = w
/ r. Given the above production function, we have MPL / MPK = K0.5 / L0.5. Thus, the tangency
condition implies K0.5 / L0.5 = . Hence, the cost minimizing labour-capital ratio is L / K =
4.
b) 6 marks. The tangency condition gives us L = 4*K. Plug this in the production function to
get K0.5 +2*K0.5 = q. Simplification yields K0.5 = q/3. Hence, K = q and L = 4*q.
Therefore, the long-run cost is C = w*L + r*K = 1*4*q + 2*q =2*q.
c) 6 marks. If K = 1, the amount of labour that is required to produce q units of output must
satisfy 10.5 + L0.5 = q. Solving for L yields L = (q – 1)2. Therefore, the short-run cost will be
CSR = w*L + r*K = 1*(q -1)2 + 2*1 = q2 – 2*q + 3.
d) 7 marks. One way to solve this question is to realize that the short run-cost will be equal to
the long-run cost when in the short-run value of K is equal to its long-run optimum. In the
short run, K is fixed at 1. From part (b), the long-run optimal value of K is K = q. Hence,
q will satisfy 1 = q. Solving this equation yields q = 3.
Question 3 25 Marks
a) 6 marks. The marginal cost is MC = y2 – 20y + 105. The average total cost is ATC = y2
/ 3 – 10y + 105 + 5 / y. The average variable cost is AVC = y – 10y + 105.
b) 6 marks. The profit-maximizing output of this firm solves p = MC. Thus, we need to
solve 105 = y2 – 20y + 105 = 105. This equation has two roots: y = 0 and y = 20. If y =
0, the firm’s profit will be – 5. If y = 20, the firm’s profit will be 3985/3. Hence, the
profit-maximizing output level is y = 20.
c) 6 marks. The shutdown price is the minimum of the average variable cost. From part
A, we have AVC = y – 10y + 105. The value of y which minimizes AVC solves
dAVC / dy = 0. Differentiation gives us 2*y / 3 – 10 = 0. Solving this equation yields y
= 15. Substitute y = 15 in AVC to get AVC = 30. Hence, the shutdown price is 30.
1.
, d) 7 marks. The lowest positive amount of output that this firm will supply is the output
that minimizes AVC. In part (c), we found that this output is y = 15.
1.
