🔬 Tutorial problems gamma

Note

This problems are designed to help you practice the concepts covered in the lectures. Not all problems may be covered in the tutorial, those left are for additional practice.

\(\gamma\).1

Formulation

Solve the following polynomial equations using the method of completing the square, and factor the left-hand side of the equation where possible.

1. \(\quad x^2 - 5x + 6 = 0\)

2. \(\quad y^2 - y - 12 = 0\)

3. \(\quad 2x^2 + 60x +800 = 0\)

4. \(\quad -\tfrac{1}{4}x^2 + \tfrac{1}{2}x + \tfrac{1}{2} = 0\)

5. \(\quad m(m-5) - 3 = 0\)

6. \(\quad 0.1 p^2 +p -2.4 = 0\)

[Sydsæter, Hammond, Strøm, and Carvajal, 2016] Exercises for Section 3.3, Question 2

Solution

1. Completing the square

\[\begin{split} \begin{array}{l} x^2 - 5x + 6 = 0 \\ x^2 - 2x\cdot\tfrac{5}{2} + \left(\tfrac{5}{2}\right)^2 + 6 - \tfrac{25}{4}= 0 \\ \left(x - \tfrac{5}{2}\right)^2 = \tfrac{25}{4} - 6 \\ x - \tfrac{5}{2} = \pm \sqrt{\tfrac{1}{4}} \\ x = \tfrac{5 \pm 1}{2} \end{array} \end{split}\]

2. Using Viète’s formulas to factor

\[\begin{split} \begin{array}{l} y^2 - y - 12 = 0 \\ (y-4)(y+3) = 0 \\ y = 4 \quad \text{or} \quad y = -3 \end{array} \end{split}\]

3. Completing the square

\[\begin{split} \begin{array}{l} 2x^2 + 60x +800 = 0 \\ x^2 + 30x + 400 = 0 \\ x^2 + 2x\cdot 15 + 225 +400 - 225 = 0 \\ (x+15)^2 = 225 - 400 \\ (x+15)^2 = - 175 \end{array} \end{split}\]

The equation has no real roots (it’s ok not to give complex solutions).

4. Completing the square

\[\begin{split} \begin{array}{l} -\tfrac{1}{4}x^2 + \tfrac{1}{2}x + \tfrac{1}{2} = 0 \\ x^2 - 2x -2 = 0 \\ x^2 - 2x +1 -2-1 = 0 \\ (x-1)^2 = 3 \\ x-1 = \pm \sqrt{3} \\ x = 1 \pm \sqrt{3} \end{array} \end{split}\]

5. Completing the square

\[\begin{split} \begin{array}{l} m(m-5) - 3 = 0 \\ m^2 -5m - 3 = 0 \\ m^2 - 2m\cdot\tfrac{5}{2} + \left(\tfrac{5}{2}\right)^2 - 3 - \tfrac{25}{4} = 0 \\ \left(m-\tfrac{5}{2}\right)^2 = 3+\tfrac{25}{4} \\ m-\tfrac{5}{2} = \pm \sqrt{\tfrac{37}{4}} \\ m = \tfrac{5}{2} \pm \sqrt{\tfrac{37}{4}} \\ m = \tfrac{5\pm \sqrt{37}}{2} \end{array} \end{split}\]

6. Completing the square

\[\begin{split} \begin{array}{l} 0.1 p^2 +p -2.4 = 0 \\ p^2 + 10p -24 = 0 \\ p^2 + 2p\cdot 5 + 25 - 24 -25 = 0 \\ (p+5)^2 = 49 \\ p+5 = \pm 7 \\ p = -5 \pm 7 \end{array} \end{split}\]

\(\gamma\).2

Formulation

Solve the following equations:

1. \(\quad x^3-4x = 0\)

2. \(\quad x^4 - 5x^2 +4 =0\)

3. \(\quad x^{-2}-2x^{-1} -15=0\)

[Sydsæter, Hammond, Strøm, and Carvajal, 2016] Exercises for Section 3.3, Question 6

Solution

1. Factorizing the LHS

\[\begin{split} \begin{array}{l} x^3-4x = 0 \\ x(x^2-4) = 0 \\ x(x-2)(x+2) = 0 \\ x_1 = 0 \quad x_2 = 2 \quad x_3 = -2 \end{array} \end{split}\]

2. Making a substitution \(y=x^2\), and using Viète’s formulas

\[\begin{split} \begin{array}{l} x^4 - 5x^2 +4 =0 \\ y^2 - 5y +4 =0 \\ (y-4)(y-1) = 0 \\ y_1 = 4 \quad y_2 = 1 \\ x = \pm 2 \quad \text{or} \quad x = \pm 1 \\ \text{Full solution: }x_1 = -2 \quad x_2 = 2 \quad x_3 = -1 \quad x_4 = 1 \end{array} \end{split}\]

3. Making a substitution \(y=1/x\), and using Viète’s formulas

\[\begin{split} \begin{array}{l} x^{-2}-2x^{-1} -15=0 \\ y^2-2y -15=0 \\ (y-5)(y+3) = 0 \\ 1/x = 5 \quad \text{or} \quad 1/x = -3 \\ \text{Full solution: }x_1 = -1/5 \quad x_2 = -1/3 \end{array} \end{split}\]

\(\gamma\).3

Formulation

The Geometric Products Company produces a product at a variable cost per unit of $2.20. The company’s fixed costs are $95,000.00 and each unit sells for $3.00. Find a function that expresses the company’s profit as a function of its output choice. What type of function is it? Compute the break-even point for the firm.

Solution

The firm’s (short-run) cost function is \(C(Q) = F + V(Q) = 95,000 + (2.20)Q\).

The firm’s revenue function is \(R(Q) = pQ = 3Q\).

The firm’s profit function is \(\Pi(Q) = R(Q) - C(Q) = 3Q - (95,000 + (2.20)Q)\) \(= \)3Q - 95,000 - (2.20)Q = (0.80)Q - 95,000$.

Clearly the firm’s profit is a linear (or an affine) function of its output.

The break-even point is given by zero profit, thus we have to solve \(\Pi(Q) = 0\). Solution is \(Q = 95,000 / 0.80 = 118,750\) units.

\(\gamma\).4

Formulation

A.

Consider the Cobb-Douglas utility function, \(U(x, y) = x^{\alpha} y^{1-\alpha}\).

We will assume that \(0 < \alpha < 1\). Assuming that \(x > 0\) and \(y > 0\), calculate \(\ln(U(x, y))\).

Why did we need to assume that \(x > 0\) and \(y > 0\)?

Calculate \(\exp(\ln(U(x, y)))\).

What, if anything, do you notice?

B.

Now Consider the Stone-Geary utility function, \(U(x, y) = (x - \beta)^{\alpha} (y - \gamma)^{1-\alpha}\).

We will assume that \(0 < \alpha < 1\), \(\beta \geq 0\), and \(\gamma \geq 0\).

Assuming that \(x > \beta\) and \(y > \gamma\), calculate \(\ln(U(x,y))\).

Why did we need to assume that \(x > \beta\) and \(y > \gamma\)?

Calculate \(\exp(\ln(U(x, y)))\).

What, if anything, do you notice?

C.

Is the Cobb-Douglas utility function a special case of the Stone-Geary utility function? Explain your answer.

Solution

A.

Consider the Cobb-Douglas utility function

\[ U(x, y) = x^\alpha y^{1-\alpha}, \]

where \(0 <\alpha< 1\) is a fixed parameter and \((x,y) \in \mathbb{R}^2_+\) is a commodity bundle.

Taking the logarithm, we have:

\[\begin{split} \begin{array}{rl} \ln(U(x,y)) =& \ln(x^\alpha y^{1-\alpha}) = \\ =& \ln(x^\alpha) + \ln(y^{1-\alpha}) = \\ =& \alpha \ln(x) + (1-\alpha) \ln(y) \end{array} \end{split}\]

We needed to assume that both \(x > 0\) and \(y > 0\) because the function \(\ln(z)\) is not defined when \(z \leq 0\). The only way to ensure that \(z = U(x, y) > 0\) is to require that either

• both \(x > 0\) and \(y > 0\), or

• both \(x < 0\) and \(y < 0\).

Since it does not make economic sense to assume that both \(x < 0\) and \(y < 0\), we need to assume that both \(x > 0\) and \(y > 0\).

If both \(x > 0\) and \(y > 0\), we have

\[\begin{split} \begin{array}{rl} \exp\big[\ln(U(x,y))\big] =& \exp\big[\alpha \ln(x) + (1-\alpha) \ln(y)\big] = \\ =& \exp\big[\ln(x^\alpha) + \ln(y^{1-\alpha}) \big] = \\ =& \exp\big[\ln(x^\alpha y^{1-\alpha}) \big] = \\ =& x^\alpha y^{1-\alpha} = U(x,y) \end{array} \end{split}\]

Note that we have recovered the original Cobb-Douglas utility function. The reason for this is that the function \(\exp(z)\) is the inverse function for the function \(\ln(z)\).

B.

Consider the Stone-Geary utility function

\[ U(x, y) = (x-\beta)^\alpha (y-\gamma)^{1-\alpha}, \]

where \(0 < \alpha < 1\), \(\beta \geq 0\), and \(\gamma > 0\) are fixed parameters, and \((x,y) \in \mathbb{R}^2_+\) is a commodity bundle.

Taking the logarithm, we have:

\[\begin{split} \begin{array}{rl} \ln(U(x,y)) =& \ln((x-\beta)^\alpha (y-\gamma)^{1-\alpha}) = \\ =& \ln((x-\beta)^\alpha) + \ln((y-\gamma)^{1-\alpha}) = \\ =& \alpha \ln(x-\beta) + (1-\alpha) \ln(y-\gamma) \end{array} \end{split}\]

We needed to assume that both \(x > \beta\) and \(y > \gamma\) because the function \(\ln(z)\) is not defined when \(z \leq 0\). The only way to ensure that \(z = U(x, y) > 0\) is to require that either

• both \(x > \beta\) and \(y > \gamma\), or

• both \(x < \beta\) and \(y < \gamma\). The point \((x,y) = (\alpha, \beta)\) is sometimes interpreted as the subsistence consumption bundle. Under this interpretation, if either \(x < \beta\) and \(y < \gamma\) or both, then the consumer is sometimes assumed to die. One way of capturing this idea is to set \(U(x, y) = -\infty\) for such cases. Under this interpretation, it might make sense to focus on the case where both \(x > \beta\) and \(y > \gamma\).

If both \(x > \beta\) and \(y > \gamma\), we have s

\[\begin{split} \begin{array}{rl} \exp\big[\ln(U(x,y))\big] =& \exp\big[\alpha \ln(x-\beta) + (1-\alpha) \ln(y-\gamma)\big] = \\ =& \exp\big[\ln((x-\beta)^\alpha) + \ln((y-\gamma)^{1-\alpha}) \big] = \\ =& \exp\big[\ln((x-\beta)^\alpha (y-\gamma)^{1-\alpha}) \big] = \\ =& (x-\beta)^\alpha (y-\gamma)^{1-\alpha} = U(x,y) \end{array} \end{split}\]

Note that we have recovered the original Stone-Geary utility function. The reason for this is that the function \(\exp(z)\) is the inverse function for the function \(\ln(z)\).

C.

Consider the Stone-Geary utility function,

\[ U(x, y) = (x-\beta)^\alpha (y-\gamma)^{1-\alpha}, \]

where \(0 < \alpha < 1\), \(\beta \geq 0\), and \(\gamma > 0\) are fixed parameters, and \((x,y) \in \mathbb{R}^2_+\) is a commodity bundle.

Suppose that we set \(\beta = 0\) and \(\gamma = 0\). Upon doing this, we obtain:

\[ U(x,y|\beta=0,\gamma=0) = (x-\beta)^\alpha(y-\gamma)^{1-\alpha} = (x-0)^\alpha(y-0)^{1-\alpha} = x^\alpha y^{1-\alpha}. \]

But this is simply the Cobb-Douglas utility function. As such, the Cobb-Douglas utility function is the special case of the Stone-Geary utility function in which both \(\beta = 0\) and \(\gamma = 0\).

\(\gamma\).5

Formulation

Consider a company that purchases cocoa in Ghana, ships it from Ghana to the United Kingdom, and sells it in the United Kingdom. The company is a monopsonist (that is, the only purchaser) in the cocoa market in Ghana. It is also a monopolist (that is, the only seller) in the cocoa market in the United Kingdom.

The price that the company receives per tonne of cocoa that it sells in the United Kingdom is given by the United Kingdom Cocoa Inverse Demand Equation, \(P_{UK} = \alpha_1 - \tfrac{1}{3}Q\), where \(Q\) is the number of tonnes of cocoa that it sells in the United Kingdom.

The price that it must pay per tonne of cocoa that it purchases in Ghana is given by the Ghana Cocoa Inverse Supply Equation, \(P_G = \alpha_2 + \tfrac{1}{6}Q\), where \(Q\) is the number of tonnes of cocoa that it purchases in Ghana.

It costs the company \(\gamma\) per tonne of cocoa to ship it from Ghana to the United Kingdom. No cocoa is lost or spoiled in this process. As such, you may assume that the quantity of cocoa purchased in Ghana, the quantity shipped from Ghana to the United Kingdom, and the quantity sold in the United Kingdom, are all identical.

1. Express the company’s revenue as a function of the quantity of cocoa it purchases, transports, and sells.

2. Express the company’s costs as a function of the quantity of cocoa it purchases, transports, and sells.

3. Express the company’s profits as a function of the quantity of cocoa it purchases, transports, and sells.

4. Assuming that \(\alpha_1 - \alpha_2 - \gamma > 0\), find the break-even quantity (or quantities) of cocoa trade for this company.

5. Identify the range (or ranges) of cocoa trade that will earn the company strictly positive and strictly negative profits.

6. How would your answer to the preceding part of this question change if \(\alpha_1 - \alpha_2 - \gamma \leq 0\)?

Solution

Part 1

The company is a monopolist in the United Kingdom cocoa market. As such, it realises that the quantity that it sells will influence the price per unit that it receives because of the downward sloping inverse demand curve for cocoa in the United Kingdom. As such, the company’s revenue is given by

\[ R(Q) = P_{UK}(Q)Q = \left(\alpha_1 - \left(\frac{1}{3}\right)Q \right)Q = \alpha_1 Q - \left(\frac{1}{3}\right) Q^2 \]

Part 2

The company’s costs consist of both the cost of purchasing the cocoa in Ghana and the cost of transporting the cocoa from Ghana to the United Kingdom. Since the company is a monopsonist in the Ghana cocoa market, it realises that the quantity of cocoa that it purchases will influence the price per unit that it must pay because of the upward sloping inverse supply curve for cocoa in Ghana. As such, the company’s revenue is given by

\[\begin{split} \begin{array}{l} C(Q) = P_{Ghana}(Q)Q + \gamma Q \\ = \left(\alpha_2 + \frac{1}{6}Q \right)Q + \gamma Q \\ = \alpha_2 Q + \frac{1}{6}Q^2 + \gamma Q \\ = (\alpha_2 + \gamma) Q + \frac{1}{6}Q^2. \end{array} \end{split}\]

Part 3

The company’s profit is simply the difference between its revenue and its costs. As such, its profit is given by

\[\begin{split} \begin{array}{l} \Pi(Q) = R(Q) - C(Q) \\ = \alpha_1 Q - \left(\frac{1}{3}\right) Q^2 - \left\{(\alpha_2 + \gamma) Q + \left(\frac{1}{6}\right) Q^2\right\} \\ = \alpha_1 Q - \left(\frac{2}{6}\right) Q^2 - (\alpha_2 + \gamma) Q - \left(\frac{1}{6}\right) Q^2 \\ = \left\{\alpha_1 - (\alpha_2 + \gamma)\right\} - \left(\frac{3}{6}\right) Q^2 \\ = (\alpha_1 - \alpha_2 - \gamma) Q - \left(\frac{1}{2}\right)Q^2. \end{array} \end{split}\]

Part 4

Note that the company’s profit is a quadratic function of \(Q\) with a negative coefficient on the quadratic term. This means that the graph of its profit function will be an “upside-down”, or hill-shaped, parabola. By definition, the company will break-even when its profit is precisely equal to zero. This requires that

\[\begin{split} \Pi(Q) = 0 \Longleftrightarrow (\alpha_1 - \alpha_2 - \gamma) Q - \left(\frac{1}{2}\right)Q^2 = 0 \\ \Longleftrightarrow -\left(\frac{1}{2}\right)Q^2 + (\alpha_1 - \alpha_2 - \gamma) Q + 0 = 0. \end{split}\]

This is a quadratic equation in \(Q\) with the coefficients \(a = -(\frac{1}{2}), b = (\alpha_1 - \alpha_2 - \gamma)\), and \(c = 0\). We can solve it by using the quadratic formula. Upon doing this, we obtain

\[\begin{split} \begin{array}{l} Q_1, Q_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ = \frac{-(\alpha_1 - \alpha_2 - \gamma) \pm \sqrt{(\alpha_1 - \alpha_2 - \gamma)^2 - 4\left(-\frac{1}{2}\right)(0)}}{2\left(-\frac{1}{2}\right)} \\ = \frac{-(\alpha_1 - \alpha_2 - \gamma) \pm \sqrt{(\alpha_1 - \alpha_2 - \gamma)^2 - 0}}{-1} \\ = (\alpha_1 - \alpha_2 - \gamma) \mp \sqrt{(\alpha_1 - \alpha_2 - \gamma)^2} \\ = (\alpha_1 - \alpha_2 - \gamma) \mp (\alpha_1 - \alpha_2 - \gamma) \\ = 0, 2(\alpha_1 - \alpha_2 - \gamma). \end{array} \end{split}\]

Part 5

Note that when \((\alpha_1 - \alpha_2 - \gamma) > 0\), we have \(Q_2 = 2(\alpha_1 - \alpha_2 - \gamma) > 0\). Since the profit function is an upside-down parabola, we know in this case that:

• \(\Pi < 0\) when \(Q < 0\) or \(Q > 2(\alpha_1 - \alpha_2 - \gamma)\),

• \(\Pi = 0\) when \(Q = 0\) or \(Q = 2(\alpha_1 - \alpha_2 - \gamma)\), and

• \(\Pi > 0\) if \(0 < Q < 2(\alpha_1 - \alpha_2 - \gamma)\).

Finally, note that \(Q < 0\) does not really make any sense in the economic context of this question. A negative choice of \(Q\) might be interpreted as the company purchasing cocoa in the United Kingdom, shipping it to Ghana and selling it in Ghana. But we are not told that the cocoa supply function in Ghana becomes the Cocoa demand function in Ghana when \(Q < 0\), nor are we told that the cocoa demand function in the United Kingdom becomes the cocoa supply function in the United Kingdom when \(Q < 0\), nor are we told that the cost of transporting cocoa from Ghana to the United Kingdom is identical to the cost of shipping it from the United Kingdom to Ghana.

Part 6

Suppose that \((\alpha_1 - \alpha_2 - \gamma) \leq 0\). Let us divide this case into two distinct sub-cases: (i) \((\alpha_1 - \alpha_2 - \gamma) = 0\), and (ii) \((\alpha_1 - \alpha_2 - \gamma) < 0\).

First, consider the case in which \((\alpha_1 - \alpha_2 - \gamma) = 0\). In this case, there is a single, repeated root, so that the unique distinct solution to the break-even equation is \(Q_1 = Q_2 = 0\). In this case, the firm can never earn a positive profit. It will earn a profit of precisely zero (that is, it will break even) when \(Q = 0\). Any other value for \(Q\) will yield a negative profit.

Now consider the case in which \((\alpha_1 - \alpha_2 - \gamma) < 0\). In this case, we once again have two distinct roots. These are the same roots that we found earlier: \(Q_1 = 0\) and \(Q_2 = 2(\alpha_1 - \alpha_2 - \gamma)\). Note that in this case, however, we have \(Q_2 = 2 (\alpha_1 - \alpha_2 - \gamma) < 0\). Since the profit function is an upside-down parabola, we know in this case that:

• \(\Pi < 0\) when \(Q < 2 (\alpha_1 - \alpha_2 - \gamma)\) or \(Q > 0\),

• \(\Pi = 0\) when \(Q = 2 (\alpha_1 - \alpha_2 - \gamma)\) or \(Q = 0\), and

• \(\Pi > 0\) if \(2 (\alpha_1 - \alpha_2 - \gamma) < Q < 0\).

Finally, note once again that \(Q < 0\) does not really make any sense in the economic context of this question. A negative choice of \(Q\) might be interpreted as the company purchasing cocoa in the United Kingdom, shipping it to Ghana and selling it in Ghana. But we are not told that the cocoa supply function in Ghana becomes the Cocoa demand function in Ghana when \(Q < 0\), nor are we told that the cocoa demand function in the United Kingdom becomes the cocoa supply function in the United Kingdom when \(Q < 0\), nor are we told that the cost of transporting cocoa from Ghana to the United Kingdom is identical to the cost of shipping it from the United Kingdom to Ghana.