🔬 Problem set beta

\(\beta\).1

Formulation

Let \(f \colon [-1, 1] \to \mathbb{R}\) be defined by \(f(x) = 1 - |x|\), where \(|x|\) is the absolute value of \(x\).

• Is the point \(x = 0\) a maximizer of \(f\) on \([-1, 1]\)?

• Is it a unique maximizer?

• Is it an interior maximizer?

• Is it stationary?

Hint

Draw a graph of function \(f\).

Solution

The point \(x=0\) is indeed a maximizer, since \(f(x) = 1 -|x| \leq 1 = f(0)\) for any \(x \in [-1, 1]\) (\(|x|=0\) if and only if \(x=0\)).

It is also a unique maximizer, since no other point is a maximizer (because \(1 -|x| < 1\) for any other \(x\)).

It is an interior maximizer since \(0\) is not an end point of \([-1, 1]\).

It is not stationary because \(f\) is not differentiable at this point (sketch the graph if you like) and hence cannot satisfy \(f'(x)=0\).

\(\beta\).2

Formulation

Consider function \(f \colon X \to \mathbb{R}\) defined by \(f(x) = \frac{1}{x} e^x\).

Find the minimizer(s) and the maximizer(s) of this function on \(X = (0, 2]\).

Follow all the required steps and explain your reasoning.

Hint

Review the algorithm for univariate optimization in the lecture notes

Solution

Following the algorithm for the univariate optimization from the lecture notes

1. Locate all stationary points

• according to the definition the stationary points are those interior points where \(f'(x) = 0\)

\[ f'(x) = \frac{d}{dx} \left( \frac{e^x}{x} \right) = \frac{e^x}{x} -\frac{1}{x^2} e^x = \frac{e^x}{x^2} \left( x - 1 \right) \]

\[ f'(x) =0 \iff x=1 \]

2. Evaluate the function at the stationary points and the boundaries, in our case only one boundary \(x=2\).

\[ f(1) = e \approx 2.718, \; f(2) = \frac{e^2}{2} \approx 3.694 \ \]

Evaluating the function at \(x=0\) is not possible because the function is not defined there! We have to be careful and investigate the behavior of the function as \(x\) approaches \(0\).

Because \(exp(x)\) is always positive, for small values of \(x\) the function takes on large numbers. The smaller \(x\) is, the larger the function becomes. This means that the function is unbounded as \(x\) approaches \(0\) (from the right). Therefore, we could move forward taking \(f(0) = \infty\).

3. Compare the values of the function at the stationary points and the boundaries to pick out the solution.

The minimizer of the function on \((0,2]\) is \(x=1\) because \(\min\{e,e^2/2,\infty\} = e\).

The maximizer of the function could be \(x=0\) because \(\max\{e,e^2/2,\infty\} = \infty\), but as the function is not defined at \(x=0\). We showed that the function grows without bound as \(x\) becomes closer and closer to 0, therefore it is impossible to find a precise \(x\) where it attains the maximum value (there is always possible to make a step towards zero to increase the function a little more). The conclusion is that there is no maximizer on \([0,2]\).

\(\beta\).3

Formulation

Find an example of a nonlinear univariate function \(f \colon D \subset \mathbb{R} \to \mathbb{R}\) that:

(a) has exactly one maximizer and one minimizer (b) has has neither a maximizer nor a minimizer (c) has an infinite number of maximizers and minimizers (d) has exactly finite number \(n\) of maximizers and \(n\) minimizers

Remember to define both the function \(f(x)\) and its domain \(D\) for each case.

Hint

First, review the relevant definitions. Then, try to draft some ideas on a piece of paper. Think of how they can be expressed in mathematical terms.

Solution

This is a creative problem which has many possible correct answers.

As always, start with the definitions—in this case definitions of maximizer and minimizer.

Here is one possible solution:

(a) any linear (affine) function \(f(x)=ax+b\), \(a \ne 0\) on any close interval \([A,B]\) has no stationary points, and therefore exactly one maximizer and one minimizer at the edges of the interval. (b) any linear (affine) function \(f(x)=ax+b\), \(a \ne 0\) on any open interval \((A,B)\) has no maximizer and no minimizer similarly to the no existence example in the lecture notes. A different idea would be to rely on positive and negative infinity in the domain \(D\), for example, let \(f(x) = \tan(\pi x)\) which takes values \(0\) on all integer points, and approaches a vertical line at every half-integer point. (c) A constant function \(f(x) = C\) should immediately come to mind. Another possibility is the cyclic trigonometric functions like \(f(x) = \sin(x)\) or \(f(x) = \cos(x)\) on the entire real line. The latter only return values between -1 and 1, and attain these two values infinitely many times. (d) This is the most tricky question, but one solution is to adjust the domain of the trig function such as \(f(x) = \cos(\pi x)\). This function attains 1 at \(x=\{...,-2,0,2,4,...\}\) and attains 0 at \(x=\{...,-3,-1,1,3,5,...\}\). Therefore, it has exactly \(n\) maximizers and \(n\) minimizers if we define \(D = [0,2n-1]\).

\(\beta\).4

Formulation

A firm uses capital and labor to produce output. When it employs \(k\) units of capital and \(\ell\) units of labor, its output is \(A k^{\alpha} \ell^{\beta}\) units, where \(A\) is a positive number, and \(\alpha + \beta < 1\).

The unit price of capital is \(r\), and the unit price of labor is \(w\); both are non-negative. The firm would like to maximize the profits taking the price \(p\) of the output as given.

The firm’s chief economist Bob presented the following formulation of the firm’s optimization problem to the CEO Alice:

\[ \text{Choose } k, \ell, w \text{ and } r \text{ to maximize } p k^{\alpha} \ell^{\beta} - w \ell - r k \quad \text{ subject to } \quad \alpha + \beta < 1 \]

Questions:

1. Is this formulation of the firm’s optimization problem correct?

• What part reflects the revenue?

• What part reflects the costs?

• What are the choice variables?

• Are there any constraints to be taken into account?

2. Right down the problem after Alice have updated the formulation.

3. Approach the problem as unconstrained maximization, and follow the steps in the lecture to find find all stationary points (solve the FOCs).

4. Write down second order partial derivatives and verify the shape conditions for the profit function.

5. What is the optimal strategy for the firm? Is the maximizer unique? Why?

Hint

Review the algorithm for optimization of bivariate functions in the lecture notes

Solution

1. The formulation is not correct. The revenue (after reincerting constant \(A\)) is \(p A k^{\alpha} \ell^{\beta}\), the costs are \(w \ell + r k\), and the choice variables are \(k\) and \(\ell\) (\(w\) and \(r\) are not chosen by the firm).
   The constraint \(\alpha + \beta < 1\) is irrelevant for the optimization problem, instead it is a constraint on the parameters for the problem to be well posed. Relevant constraints on the optimization problem are \(k>0\) and \(\ell>0\), they can be first ignored and checked after we solve the unconstrained version of the problem.

2. The correct formulation is (\(A, p, \alpha, \beta, w, r\) are parameters and should be fixed/found out before the firm solves the optimization problem)

\[ p A k^{\alpha} \ell^{\beta} - w \ell - r k \rightarrow \max_{k, \ell} \]

\[ \quad \text{subject to} \quad k \ge 0, \ell \ge 0 \]

3. See lecture notes

4. See lecture notes

5. Optimal strategy \(k^*, \ell^*\) are given in the lecture notes. The maximizer is unique because the objective function is strictly concave when \(\alpha+\beta < 1\).

Proof:

We check second order conditions for strict concavity.

What we need: for any \(k, \ell > 0\)

1. \(\pi_{11}(k, \ell) < 0\)

2. \(\pi_{11}(k, \ell) \, \pi_{22}(k, \ell) > \pi_{12}(k, \ell)^2\)

The second order derivatives are

\[ \pi_{11}(k,\ell) = (\alpha-1)\alpha pA k^{\alpha-2} \ell^\beta \]

\[ \pi_{22}(k,\ell) = (\beta-1)\beta pA k^{\alpha} \ell^{\beta-2} \]

\[ \pi_{12}(k, \ell) = \alpha \beta pA k^{\alpha-1} \ell^{\beta-1}. \]

Since \(\alpha+\beta<1\) and \(\alpha, \beta \geq 0\), we have \(\alpha-1<0\), which implies \(\pi_{11}(k,\ell)<0\) for all \(k, \ell >0\).
Moreover, the second order differentials imply

\[ \pi_{11}(k, \ell) \, \pi_{22}(k, \ell) = (\alpha-1)(\beta-1)\alpha \beta p^2 A^2 k^{2\alpha-2} \ell^{2\beta-2} \]

\[ (\pi_{22}(k, \ell))^2 = \alpha^2 \beta^2 p^2 A^2 k^{2\alpha-2} \ell^{2\beta-2}. \]

Assuming that all parameters and variables are positive. Then, we obtain \(\pi_{11}(k, \ell) \, \pi_{22}(k, \ell) > \pi_{12}(k, \ell)^2\) if and only if \((\alpha-1)(\beta-1) > \alpha \beta\) if and only if \(1 > \alpha + \beta\).