🔬 Tutorial problems iota

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. The symbol 🍹 indicates additional problems.

\(\iota\).1

Formulation

Cobb-Douglas Preferences Specific Example:

Consider the utility function \(U(x, y)=x^{0.5} y^{0.5}\) that is defined on the consumption set \(\mathbb{R}_{+}^{2}\).

(a) Find the equation of the indifference curve that corresponds to \(U=40\).

(b) What is the slope of the indifference curve for \(U=40\) for any given value of \(x\)?

(c) What is the equation of an arbitrary indifference curve for this utility function?

(d) What is the slope of an arbitrary indifference curve for this utility function at any given value of \(x\)?

Solution

2(a)

The utility function is

\[ U(x, y)=U(x, y)=x^{0.5} y^{0.5}=\sqrt{x} \sqrt{y}=\sqrt{x y} \]

The indifference curve corresponding to an amount of utility equal to 40 “utils” is given by the condition

\[ U(x, y)=40 \Longleftrightarrow \sqrt{x y}=40 \Longleftrightarrow x y=1,600 \Longleftrightarrow y=\frac{1,600}{x} \]

Thus the equation of the indifference curve corresponding to an amount of utility equal to 40 “utils” is

\[ y=\frac{1,600}{x} \]

Note that the graph of this indifference curve will be a rectangular hyperbola.

2(b)

The slope of the indifference curve corresponding to an amount of utility equal to 40 “utils” is given by

\[ \frac{d y}{d x}=\frac{d}{d x}\left(\frac{1,600}{x}\right)=\frac{d}{d x} 1,600 x^{-1}=-1,600 x^{-2}=-\left(\frac{1,600}{x^{2}}\right) \]

We can show that the marginal rate of substitution of \(x\) for \(y\) for this utility function is given by

\[ M R S_{x, y}(x, y)=\frac{y}{x} \]

(Exercise: Show this.) At first glance, it might appear that

\[ \frac{d y}{d x} \neq-M R S_{x, y}(x, y) \]

in this case (when \(U=40\) ). Oh ye of little faith! Recall that the equation for the indifference curve corresponding to a utility level of forty “utils” was

\[ y=\frac{1,600}{x} \]

Upon substituting this into the marginal rate of substitution expression, we obtain

\[ M R S_{x, y}(x, y)=\frac{y}{x}=\frac{\left(\frac{1,600}{x}\right)}{x}=\frac{1,600}{x^{2}} \]

so that we do indeed have

\[ \frac{d y}{d x}=\frac{1,600}{x^{2}}=-M R S_{x, y}(x, y) \]

2(c)

The utility function is

\[ U(x, y)=U(x, y)=x^{0.5} y^{0.5}=\sqrt{x} \sqrt{y}=\sqrt{x y} \]

The indifference curve corresponding to an arbitrary, but given, amount of utility, say \(k>0\), is given by the condition

\[ U(x, y)=k \Longleftrightarrow \sqrt{x y}=k \Longleftrightarrow x y=k^{2} \Longleftrightarrow y=\frac{k^{2}}{x} \]

Thus the equation of the indifference curve corresponding to a utility level of \(k>0\) is

\[ y=\frac{k^{2}}{x} \]

Note that the graph of any such indifference curve (when the utility level is \(k>0\) ) will be a rectangular hyperbola.

2(d)

The slope of the indifference curve corresponding to a utility level of \(k>0\) is given by

\[ \frac{d y}{d x}=\frac{d}{d x}\left(\frac{k^{2}}{x}\right)=\frac{d}{d x} k^{2} x^{-1}=-k^{2} x^{-2}=-\left(\frac{k^{2}}{x^{2}}\right) \]

We can show that the marginal rate of substitution of \(x\) for \(y\) for this utility function is given by

\[ M R S_{x, y}(x, y)=\frac{y}{x} \]

(Exercise: Show this.) At first glance, it might appear that

\[ \frac{d y}{d x} \neq-M R S_{x, y}(x, y) \]

in this case (when \(U=k>0\) ). Oh ye of little faith! Recall that the equation for the indifference curve corresponding to a utility level of \(k>0\) was

\[ y=\frac{k^{2}}{x} \]

Upon substituting this into the marginal rate of substitution expression, we obtain

\[ M R S_{x, y}(x, y)=\frac{y}{x}=\frac{\left(\frac{k^{2}}{x}\right)}{x}=\frac{k^{2}}{x^{2}} \]

so that we do indeed have

\[ \frac{d y}{d x}=\frac{k^{2}}{x^{2}}=-M R S_{x, y}(x, y) \]

\(\iota\).2

Formulation

Marginal Rates of Substitution:

Calculate the marginal rate of substitution for an arbitrary commodity bundle of the form \((x, y)>>\) \((0,0)\) (that is, where \(x>0\) and \(y>0\) ) for each of the following utility functions.

(a) Quasi-Linear Preferences Example 1: \(U(x, y)=x+\sqrt{y}\).

(b) Quasi-Linear Preferences Example 2: \(U(x, y)=x+\ln (y)\).

(c) Stone-Geary Preferences: \(U(x, y)=\left(x-x_{0}\right)^{\alpha}\left(y-y_{0}\right)^{1-\alpha}\), where \(x_{0}>0, y_{0}>0\), and \(\alpha \in(0,1)\) are fixed parameters.

(d) Constant-Elasticity-of-Substitution (CES) Preferences: \(U(x, y)=\left(\alpha x^{\rho}+\beta y^{\rho}\right)^{\frac{1}{\rho}}\), where \(x_{0}>0, y_{0}>0\), and \(\alpha \in(0,1)\) are fixed parameters.

Solution

3(a)

The utility function is

\[ U(x, y)=x+\sqrt{y}=x+y^{\frac{1}{2}} \]

The marginal utility of commodity \(x\) is

\[ M U_{x}(x, y)=\frac{\partial U(x, y)}{\partial x}=1 \]

The marginal utility of commodity \(y\) is

\[ M U_{y}(x, y)=\frac{\partial U(x, y)}{\partial y}=\left(\frac{1}{2}\right) y^{-\left(\frac{1}{2}\right)}=\frac{1}{2 \sqrt{y}} \]

The marginal rate of substitution of commodity \(x\) for commodity \(y\) is

\[ M R S_{x, y}(x, y)=\frac{M U_{x}(x, y)}{M U_{y}(x, y)}=\frac{1}{\left(\frac{1}{2 \sqrt{y}}\right)}=2 \sqrt{y} \]

3(b)

The utility function is

\[ U(x, y)=x+\ln y \]

The marginal utility of commodity \(x\) is

\[ M U_{x}(x, y)=\frac{\partial U(x, y)}{\partial x}=1 \]

The marginal utility of commodity \(y\) is

\[ M U_{y}(x, y)=\frac{\partial U(x, y)}{\partial y}=\frac{1}{y} \]

The marginal rate of substitution of commodity \(x\) for commodity \(y\) is

\[ M R S_{x, y}(x, y)=\frac{M U_{x}(x, y)}{M U_{y}(x, y)}=\frac{1}{\left(\frac{1}{y}\right)}=y \]

3(c)

The utility function is

\[ U(x, y)=\left(x-x_{0}\right)^{\alpha}\left(y-y_{0}\right)^{1-\alpha} \]

The marginal utility of commodity \(x\) is

\[ M U_{x}(x, y)=\frac{\partial U(x, y)}{\partial x}=\alpha\left(x-x_{0}\right)^{\alpha-1}\left(y-y_{0}\right)^{1-\alpha} \]

The marginal utility of commodity \(y\) is

\[\begin{split} \begin{array}{ll} M U_{y}(x, y) &= \frac{\partial U(x, y)}{\partial y}=(1-\alpha)\left(x-x_{0}\right)^{\alpha}\left(y-y_{0}\right)^{1-\alpha-1} \\ & =(1-\alpha)\left(x-x_{0}\right)^{\alpha}\left(y-y_{0}\right)^{-\alpha} \end{array} \end{split}\]

The marginal rate of substitution of commodity \(x\) for commodity \(y\) is

\[\begin{split} \begin{array}{c} M R S_{x, y}(x, y)=\frac{M U_{x}(x, y)}{M U_{y}(x, y)}=\frac{\alpha\left(x-x_{0}\right)^{\alpha-1}\left(y-y_{0}\right)^{1-\alpha}}{(1-\alpha)\left(x-x_{0}\right)^{\alpha}\left(y-y_{0}\right)^{-\alpha}} \\ =\frac{\alpha\left(y-y_{0}\right)}{(1-\alpha)\left(x-x_{0}\right)} \end{array} \end{split}\]

3(d)

The utility function is

\[ U(x, y)=\left(\alpha x^{\rho}+\beta y^{\rho}\right)^{\frac{1}{\rho}} \]

The marginal utility of commodity \(x\) is

\[\begin{split} \begin{array}{c} M U_{x}(x, y)=\frac{\partial U(x, y)}{\partial x}=\left(\frac{1}{\rho}\right)\left(\alpha x^{\rho}+\beta y^{\rho}\right)^{\left(\frac{1}{\rho}\right)-1} \rho \alpha x^{\rho-1} \\ =\alpha x^{\rho-1}\left(\alpha x^{\rho}+\beta y^{\rho}\right)^{\left(\frac{1}{\rho}\right)-1}=\alpha x^{\rho-1}\left(\alpha x^{\rho}+\beta y^{\rho}\right)^{\frac{1-\rho}{\rho}} \end{array} \end{split}\]

The marginal utility of commodity \(y\) is

\[\begin{split} \begin{array}{ll} M U_{y}(x, y)&=\frac{\partial U(x, y)}{\partial y} \\ &=\left(\frac{1}{\rho}\right)\left(\alpha x^{\rho}+\beta y^{\rho}\right)^{\left(\frac{1}{\rho}\right)-1} \rho \beta y^{\rho-1} \\ & =\beta y^{\rho-1}\left(\alpha x^{\rho}+\beta y^{\rho}\right)^{\left(\frac{1}{\rho}\right)-1}\\ &=\beta y^{\rho-1}\left(\alpha x^{\rho}+\beta y^{\rho}\right)^{\frac{1-\rho}{\rho}}\\ \end{array} \end{split}\]

The marginal rate of substitution of commodity \(x\) for commodity \(y\) is

\[\begin{split} \begin{array}{c} M R S_{x, y}(x, y)=\frac{M U_{x}(x, y)}{M U_{y}(x, y)}=\frac{\alpha x^{\rho-1}\left(\alpha x^{\rho}+\beta y^{\rho}\right)^{\frac{1-\rho}{\rho}}}{\beta y^{\rho-1}\left(\alpha x^{\rho}+\beta y^{\rho}\right)^{\frac{1-\rho}{\rho}}} \\ =\frac{\alpha x^{\rho-1}}{\beta y^{\rho-1}}=\frac{\alpha y^{1-\rho}}{\beta x^{1-\rho}}=\left(\frac{\alpha}{\beta}\right)\left(\frac{y}{x}\right)^{1-\rho} \end{array} \end{split}\]

\(\iota\).3

Formulation

Let \(U\left(q_{1}, q_{2}\right)\) be a utility function that represents a particular individual’s preferences over bundles of strictly positive amounts of each of two commodities. Suppose that this utility function is at least twice continuously differentiable. The indifference curve corresponding to utility level \(\hat{U}\) for this individual is a graphical representation of the level-set for this utility function that corresponds to utility-level \(\hat{U}\). It is formally defined to be

\[ U^{0}\left(q_{1}, q_{2}, \hat{U}\right)=\left\{\left(q_{1}, q_{2} \in \mathbb{R}_{++}^{2}: U\left(q_{1}, q_{2}\right)=\hat{U}\right\}\right. \]

The equation that describes this curve takes the form \(q_{2}=f\left(q_{1}\right)\), for some function \(f\). The function \(f\) is implicitly defined by the equation

\[ U\left(q_{1}, q_{2}\right)=\hat{U} \]

Use the equation \(U\left(q_{1}, q_{2}\right)=\hat{U}\) to show that the slope of the above indifference curve is given by

\[ \left.\frac{d q_{2}}{d q_{1}}\right|_{U=\hat{U}}=-M R S_{1,2}\left(q_{1}, q_{2}\right) \]

where MRS stands for the marginal rate of substitution, and is defined to be

\[ M R S_{1,2}\left(q_{1}, q_{2}\right)=\frac{\left(\frac{\partial U\left(q_{1}, q_{2}\right)}{\partial q_{1}}\right)}{\left(\frac{\partial U\left(q_{1}, q_{2}\right)}{\partial q_{2}}\right)} \]

Solution

• An indifference curve is a level-set for the utility function. Every consumption bundle on an indifference curve must generate the same level of utility. Thus the equation that describes a representative indifference curve is \(U\left(q_{1}, q_{2}\right)=k\), where \(k\) is a constant.

• Along any indifference curve, utility is constant. Thus if we imagine varying \(q_{1}\) and \(q_{2}\) while stating on an indifference curve, we must have no change in the utility level \((d k=0)\).

• A first-order differential approximation to the change in utility that is induced by varying \(q_{1}\) and \(q_{2}\) is given by \(d U \approx\left(\frac{\partial U}{\partial q_{1}}\right) d q_{1}+\left(\frac{\partial U}{\partial q_{2}}\right) d q_{2}=\) \(M U_{1} d q_{1}+M U_{2} d q_{2}\).

• Thus at any point on an indifference curve, we must have \(M U_{1} d q_{1}+\) \(M U_{2} d q_{2} \approx 0\).

• Rearranging this equation, we obtain

\[ \text { Slope of IC }=\left(\frac{d q_{2}}{d q_{1}}\right)_{d U=0} \approx-\left(\frac{M U_{1}}{M U_{2}}\right)=-M R S_{1,2} \]

\(\iota\).4

Formulation

Find the degree of homogeneity, if there is one, for each of the following functions:

(a) \(f(x,y,z) = 3x+4y-3z\)

(b) \(g(x,y,z) = 3x+4y-2z-2\)

(c) \(h(x,y,z) = \frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{x+y+z}\)

(d) \(G(x,y) = \sqrt{xy} \ln \left( \frac{x^2+y^2}{xy} \right)\)

(e) \(H(x,y) = \ln x + \ln y\)

(f) \(p(x_1,\dots,x_n) = \sum_{i=1}^{n} x_i^n\)

This question comes from Sydsæter, Hammond, Strøm, and Carvajal (2016) (Section 12.7, Problem 1)

Solution

To investigate whether a function is homogenous of any degree, we can check if the condition in the definition of homogenous function is satisfied.

(a)

\[ f(\lambda x,\lambda y,\lambda z) = 3\lambda x+4\lambda y-3\lambda z = \lambda (3x+4y-3z) = \lambda f(x,y,z) \]

Hence, \(f(x,y,z)\) is homogenous of degree one

(b)

\[ g(\lambda x,\lambda y,\lambda z) = \lambda(3x+4y-2z)-2 =\lambda g(x,y,z)-2 \ne \lambda^r g(x,y,z) \forall r \]

Hence, \(g(x,y,z)\) is not homogenous of any degree

(c)

\[ h(\lambda x,\lambda y,\lambda z) = \frac{\sqrt{\lambda x}+\sqrt{\lambda y}+\sqrt{\lambda z}}{\lambda x+\lambda y+\lambda z} = \left(\frac{\sqrt{\lambda}}{\lambda}\right) \left(\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{x+y+z}\right) = \lambda^{-0.5} h(x,y,z) \]

Hence, \(f(x,y,z)\) is homogenous of degree \(-\frac{1}{2}\)

(d)

\[ G(\lambda x,\lambda y) = \sqrt{\lambda^2 xy} \ln \left( \frac{\lambda^2( x^2+y^2)}{\lambda^2 xy} \right) = \lambda G(x,y) \]

Hence, \(f(x,y,z)\) is homogenous of degree one

(e)

\[ H(\lambda x,\lambda y) = \ln (\lambda x) + \ln (\lambda y) = \ln \lambda + \ln x + \ln \lambda + \ln y = H(x,y) + 2 \ln \lambda \]

Hence, \(H(x,y)\) is not homogenous of any degree

(f)

\[ p(\lambda x_1,\dots,\lambda x_n) = \sum_{i=1}^{n} (\lambda x_i)^n = \lambda^n \sum_{i=1}^{n} x_i^n = \lambda^n p(x_1,\dots,x_n) \]

Hence, \(f(x,y,z)\) is homogenous of degree \(n\)