Chapter 2 Solutions

  1. Suppose \(U(x,y)=4x^2+3y^2\)

    1. Calculate \(\frac{\partial U}{\partial x}\), and \(\frac{\partial U}{\partial y}\)

      \(\frac{\partial U}{\partial x}=8x\), \(\frac{\partial U}{\partial y}=6y\)

    2. Evaluate these partial derivatives ad \(x=1\) and \(y=2\)

      \(\frac{\partial U}{\partial x}\vert_{(1,2)}=8\), \(\frac{\partial U}{\partial y}\vert_{(1,2)}=12\)

    3. Write the total differential for \(U\).

      \(dU=\frac{\partial U}{\partial x}*dx + \frac{\partial U}{\partial y}*dy\) \(dU=8x*dx + 6y*dy\)

    4. Calculate \(\frac{dy}{dx}\) for \(dU=0\); that is, what is the implied trade-off between \(x\) and \(y\) holding \(U\) constant.

      \(0=8xdx + 6ydy\)
      \(-8xdx = 6ydy\)
      \(-\frac{4}{3}\frac{x}{y} = \frac{dy}{dx}\)

      A note about Markdown and Latex. If you want to align the equations at the equal sign, you need a bit more straight Latex.

        $$
        \begin{align}
                                0 & = 8xdx + 6ydy  
                            -8xdx & = 6ydy  
          -\frac{4}{3}\frac{x}{y} & = \frac{dy}{dx}
        \end{align}
        $$
\[\begin{align} 0 & = 8xdx + 6ydy \\ -8xdx & = 6ydy \\ -\frac{4}{3}\frac{x}{y} & = \frac{dy}{dx}\\ \end{align}\]

  1. Show \(U=16\) when \(x = 1\) and \(y = 2\).
    \(U(1,2) = 4*1^2 + 3*2^2 = 4 + 12 = 16\)

  2. In what ratio must \(x\) and \(y\) change to hold \(U\) constant at 16 for movements away from \(x=1\).
    \(\frac{dy}{dx}=-\frac{4}{3}*\frac{1}{2} = - \frac{2}{3}\)

  3. More generally, what is the slope of the \(U=16\) contour line for this function? What is the slope of that line?

    Contour line is an ellipse.
    \(16 = 4x^2 + 3y^2 \Rightarrow y = \pm \frac{2}{\sqrt 3}\sqrt{4-x^2}\)

    Slope
    \(\frac{dy}{dx} = \frac{1}{\sqrt{3}}(4-x^2)^{-\frac{1}{2}}(-2x)\)

2.2 Suppose a firm’s total revenues depend on the amount produced \((q)\) according to the function

\[R = 70q - q^2\]

Total costs also depend on \(q\):

\[C=q^2 + 30q + 100\]

  1. What level of output should the firm produce in order to maximize profits \((R-C)\)? What will profits be?

    \[\pi = 70q - q^2-30q^2-100\]

    Need to maximize profit with respect to \(q\).

    \[\frac{d\pi}{dq} = 70 - 4q - 30 = 40 - q\]

    Set first order condition equal to 0.

    \[40 - 4q = 0\]

    \[\frac{40}{4} = q\]

    \[10 = q\]

    Profit at \(\pi\) max

    \[\pi\vert_{q=10} = 700 - 100 - 100 - 300 - 100\]

  2. Show that the second-order conditions for a maximum are satisfied at the output level found in part (a)?

    \[\frac{d^2\pi}{dq^2} = -4 < 0 \]

    Second derivative is negative, so the function is concave, and therefore local optimum is a maximum.

  3. Does the solution calculated here obey the “marginal revenue equals marginal cost” rule?

    Yes. \(MR = 70 - 2q\) and \(MC = 2q + 30\).

    \[MR = MC \Rightarrow 70 - 2q = 2q + 30 \Rightarrow 40 - 4q = 0\]

    Which is the first order condition we solved to get our solution.

2.3 Suppose that \(f(x,y)=xy\). Find the maximum value for \(f\) if \(x\) and \(y\) are constrained to sum to 1. Solve this problem two ways: by substitution and by using the Lagrangian method.

  1. By substitution

Substitute \(y = 1-x\) into the objective function so that \(f(x) = x(1-x) = x - x^2\). Then maximize \(f\) with respect to \(x\).

FOC: \(1 - 2x = 0\)
\(\Rightarrow 1 = 2x\)
\(\Rightarrow \frac{1}{2} = x\)

Then \(y = 1 - \frac{1}{2} = \frac{1}{2}\)
and \(f(\frac{1}{2},\frac{1}{2}) = \frac{1}{4}\)

  1. By the Lagrangian Method

\[\max_{x, y, \lambda} xy + \lambda[1-x-y]\]

FOCs:

\(f_1: y - \lambda = 0\)
\(f_2: x- \lambda = 0\)
\(\lambda: 1 - x - y = 0\)

Using \(f_1\) and \(f_2\) by putting \(\lambda\) on one side and taking the ratio of the first order conditions.

\(y = \lambda\)
\(x = \lambda\)

\(\Rightarrow y = x\)

Now taking \(y = x\) and putting it into the FOC for \(\lambda\) we get
\(1 - x - x = 0 \Rightarrow \frac{1}{2} = y\) and \(\frac{1}{2} = x\) and \(\frac{1}{2} = \lambda\)

Finally, \(f(\frac{1}{2},\frac{1}{2}) = \frac{1}{4}\)

2.4 The dual problem to the one described in Problem 2.3 is

\[\textbf{minimize } x + y\] \[\textbf{subject to } xy = 0.25\]

Solve this problem using the Lagrangian technique. Then compare the value you get for the Lagrangian multiplier to the value you got in 2.3. Explain the relationship between the two solutions.

\[\min_{x, y, \lambda} x+y + \lambda[0.25 - xy]\]

FOC:

\(f_1: 1 - \lambda y = 0\)
\(f_2: 1 - \lambda x = 0\)
\(f_{\lambda}: 0.25 - xy = 0\)

Combine the first two equations as follows.

\(1 = \lambda y\)
\(1 = \lambda x\)

Take the ratio of the left hand side and the ratio of the right hand side and obtain

\(1 = \frac{y}{x}\) or \(x = y\). Then take this and substitute into the FOC of \(\lambda\).

\(0.25 = x^2\) \(\Rightarrow x = \frac{1}{2}\)

Then, since \(x = y\), \(y = \frac{1}{2}\), and taking both these results and plugging into the \(f_1\) FOC we find that \(\lambda = 2\).

Comparing this solution with the solution from 2.3 we find that \(x\) and \(y\) are the same. However, \(\lambda\) of the problem in 2.4 is \(\frac{1}{\lambda_{2.3}}\).

The intuition of this is that \(\lambda\), often referred to as the shadow value, can be understood as

\[\lambda = \frac{\text{Marginal Benefit }x_i}{\text{Marginal Cost }x_i}\] for all \(i\).

For the when two problems are ‘dual’ the marginal benefit in one becomes the marginal cost in the other.

2.5 The height of a ball that is thrown straight up with a certain force is a function of the time (t) from which it is released given by \(f(t) = -0.5gt^2 + 40t\) (Where \(g\) is a constant determined by gravity).

  1. How does the value of \(t\) at which the height of the ball is at a maximum depend on the parameter \(g\)?

When we maximize \(f\) with respect to \(t\), we will find that it is a function of \(g\).

\(\frac{df}{dt} = -gt + 40 = 0\)
\(t^*(g) = \frac{40}{g}\)

  1. Use your answer to part (a) to describe how the maximum height changes as the parameter \(g\) changes.

Plug optimal \(t\) into \(f(.)\). \(f(g) = -\frac{1}{2}g\frac{40}{g}^2 + 40\frac{40}{g} = \frac{800}{g}\).

How does the max change as \(g\) changes? Take the derivative.

\(f'(g) = -\frac{800}{g^2}\).

  1. By the envelope theorem, when we differentiate the value function \(f(t^*(g),g)\) with respect to \(g\), we will only get a non-zero contribution where the \(g\) directly enters the function. All contributions that come indirectly through \(t^*(g)\) are zero.

\[f(t^*(g), g) = -0.5g(t^*(g))^2 + 40(t^*(g))\]

So,

\[\begin{align*} \frac{\partial f(t^*(g), g)}{\partial g} &= -0.5(t^*(g))^2 \\ &= -0.5\left(\frac{40}{g}\right)^2 \\ &= -\frac{800}{g^2} \end{align*}\]

The same as in part b.

  1. Using the function in levels \(f(32.1) - f(32) = \frac{800}{32.1} - \frac{800}{32} = -0.078\). However, we can also calculate it with the derivative. We found \(f'(g) = -\frac{800}{g^2}\) so \(f'(32)=-0.78\). and this is understood to be the per unit change in f as g changes. In this case we are not changing g by 1 we are changing g by 0.1. If we scale the derivative in this way, we get roughly the same answer. \(0.1*f'(32)=-0.78*0.1 = -0.078\).

2.6

A simple way to model the construction of an oil tanker is to start with a large rectangular sheet of steel that is x feet wide and 3x feet long. Now cut a smaller square that is \(t\) feet on a side out of each corner of the larger sheet and fold up and weld the sides of the steel sheet to make a tray like structure with no top.

[Figure Here]

  1. show that the volume of oil that can be held by this tray is given by: \(V = t(x - 2t)(3x - 2t) = 3tx^2 - 8t^2x + 4t^3\).

Since volume is height X width X length, we calculate each of these and multiply.

Length: \(3x - 2t\) Width: \(x - 2t\) height: \(t\)

So, \(v = t(x - 2t)(3x - 2t) = 3tx^2 - 8t^2x + 4t^3\)

  1. How should t be chosen so as to maximize \(V\) for any given value of x?

\[\max_{t} 3tx^2 - 8t^2x + 4t^3\]

FOC: \(3x^2 - 16tx + 12t^2 = 0\)

Solve with the quadratic formula for t.

\(t = \frac{16x \pm \sqrt{(-16x)^2 - 4*12*3*x^2}}{2*12}\)
\(t = x(\frac{4 \pm \sqrt{7}}{6})\)

  1. No, there is not a value of x that maximizes volume. If you increase x, volume will increase as well.

  2. Now that steel is constrained so that \(1,000,000 = 3x^2 - 4t^2\), we still know that \(t\) and \(x\) have to be related according to the solution in b, but the \(t\) and \(x\) are now bound to the constraint. We can visualize the solution as a guide. The constraint is an ellipse, and only the portions with positive \(t\) and \(x\) are relevant to this problem.

[Figure Here]

So, if we plug the expression for \(t\) that we found in b. into the constraint, we can solve. Note that since both \(4 \pm \sqrt{y}\) are positive we will get two solutions here.

\[1,000,000 = 3x^2 - 4\left[x\left(\frac{4 \pm \sqrt{7}}{6}\right)\right]^2\]

\[1,000,000 = x^2\left[ 3 - 4\left(\frac{4 \pm \sqrt{7}}{6} \right)\right]\]

\[\Rightarrow x = 724.08\] and plugging this into the relationship from b. \[t = 801.92\]

or…

\[\Rightarrow x = 598.02\] and plugging this into the relationship from b. \[t = 134.98\] when using the - of the \(\pm\).

However, if you plug the first into the formula for \(V\), you get a negative number; that must be a local minimum. So, the solution to the constrained problem is \(x = 598.02\) and \(t = 134.98\).

2.7 Consider the following constrained maximization problem:

\[\textbf{maximize } y = x_1 + 5\ln{x_2}\] \[\textbf{subject to } k - x_1 - x_2 = 0\]

where \(k\) is a constant.

  1. Show that if \(k = 10\) this problem can be solved as one involving only equality constraints.

\(k = 10 \Rightarrow\)

\[\max x_1 + 5\ln{10-x_1}\] by setting \(x_2 = 10 - x_1\).

FOC: \(1 - \frac{5}{10 - x_1} = 0\)

\(\Rightarrow x_1 = 5\) and \(x_2 = 10\)

  1. If \(k = 4\) then the FOC becomes \(1 - \frac{5}{4 - x_1} = 0\) and this solves to \(x_1 = -1\) and \(x_2 = 5\).

  2. If the \(x's\) must be non-negative, then \(x_1 = 0\) and the constraint defines the \(x_2\). In this case, \(x_2 = 4\).

  3. If \(k = 20\) then the FOC becomes \(1 - \frac{5}{20 - x_1} = 0\) and the solution is \(x_1 = 15\) and \(x_2 = 5\).

2.8 Suppose a firm has a marginal cost function given by \(MC(q) = q+1\).

  1. What is the firm’s total cost function,? Explain why total costs are known only up to a constant of integration, which represents fixed costs.

You find total cost by integrating the marginal cost, but of course, you can only recover this up to an unknown constant. Since the portion of the total cost function that is constant is the fixed cost - by definition - this is the fixed cost.

\[\int q + 1 dq\]

  1. Firms produce where price, \(p\), equals \(MC(q)\). How much will the firm produce when \(p = 15\)? Now assume that the firm just breaks even. What are the fixed costs, \(FC\)?

\(15 = q + 1\) \(\Rightarrow q = 14\).

If just breaking even, then \(15*14 - (0.5q^2 + q + FC) = 0\)

\(\Rightarrow FC = 98\)

  1. How much will profits for this firm increase if price increases to $20?

First, \(20 = q + 1\) so \(q = 19\). Then profits are given by \(20*19 - (0.5*19^2 + 19 +98) = 82.50\).

  1. Show that if we continue to assume profit maximization, then this firm’s profit can be expressed solely as a function of the price it receives.

Then we may as well substitute \(q = P - 1\) from the \(P = MC(q)\) condition.

Then we have \(\pi(P) = P(P-1) - (0.5(P-1)^2 + (P-1) + 98)\)

  1. Show that the increase in profits from \(P=15\) to \(P=20\) can be calculated in two ways: (i) directly from the equation derived in part d; and by integrating the inverse marginal cost function \(\left[ MC^{-1}(P) = P-1\right]\).

(i). Clearly \(\pi(20) - \pi(15) = \$82.50\) yields the change in profit.

(ii). \(\int_{15}^{20} P-1dP = (0.5P^2 - P)\vert_{15}^{20} = \$82.50\)