Optimization Problems II: Business and Economic Optimization Problems

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In here, we discussed several examples of optimizations problems, mostly geometric optimization problems. In this note, we study business and economic optimization problems. Let us begin with the following example.

Example. Suppose that the price and demand for a particular luxury automobile are related by the demand equation $p+10x=200,000$, where $p$ is the price per car in dollars and $x$ is the number of cars that will be purchased at that price. What price should be charged per car if the total revenue is to be maximized?

Solution. Recall that the total revenue $R$ is given by $R=xp$, where $p$ is the price per item and $x$ is the number of items sold. In terms of $x$, the revenue is written as
$$R(x)=200,000x-10x^2$$
$R(x)=200,000-20x$ and set it equal to 0, we find the critical point $x=10,000$. Since $R^{\prime\prime}(x)=-20<0$, $R(10,000)=100,000,0000$, i.e. a billion dollars. We don’t actually calculus to see this. $R(x)$ is a quadratic function with negative leading coefficient, so it assumes as maximum at the $x$-coordinate of its vertex, $x=-\frac{b}{2a}=-\frac{200,000}{2\cdot 10}=10,000$. To answer the question, the price at which the total revenue $R$ is maximized is
$$p=-10(10,000)+200,000=100,000\ \mathrm{dollars}$$
As you might have noticed by now, it could have been shorter if we wrote $R$ in terms of the price $p$ since the question is about the price at which the revenue is maximized. In terms of $p$, $R$ is written as
$$R(p)=-\frac{p^2}{10}+20,000p$$
Either by solving $R'(p)=-\frac{p}{10}+20,000=0$ or by $p=-\frac{b}{2a}=\frac{20,000}{\frac{2}{10}}$, we find $p=\$ 100,000$.

Example. Suppose that the cost, in dollars, of producing $x$ hundred bicycles is given by $C(x)=x^2-2x+4900$. What is the minimum cost?

Solution. Either by solving $C'(x)=2x-2=0$ or by $x=-\frac{b}{2a}=-\frac{-2}{2\cdot 1}$, we find that $C(x)$ assumes the minimum at $x=1$, i.e. the cost of production is the minimum when 100 bicycles are produced. The minimum cost is $C(1)=4899$ dollars.

Example. In the preceding example, find the minimum average cost.

Solution. Recall that the average cost $\bar C(x)$ is given by $\bar C(x)=\frac{C(x)}{x}$, so we have
$$\bar C(x)=x-2+\frac{4900}{x}$$
Setting $C'(x)=1-\frac{4900}{x^2}$ equal to 0, we find $x=70$. Since $C^{\prime\prime}(x)=\frac{9800}{x^3}>0$ when $x=70$, $\bar C(70)=138$ (in dollars) is the minimum average cost.

Here is an interesting theorem from economics.

Theorem. The average cost is minimized at a level of production at which marginal cost equals average cost, i.e. when
$$C'(x)=\bar C(x)$$

Proof. Since $\bar C(x)=\frac{C(x)}{x}$, we obtain by the quotient rule
$$\bar C'(x)=\frac{xC'(x)-C(x)}{x^2}$$
Setting $\bar C'(x)=0$, we have
$$xC'(x)-C(x)=0,$$
that is
$$C'(x)=\frac{C(x)}{x}=\bar C(x)$$

The preceding example can be quickly answered using this Theorem. Setting $C'(x)=\bar C(x)$, we have
$$2x-2=x-2+\frac{4900}{x}$$
Simplifying this we obtain
$$x^2=4900$$
Hence, $x=70$ as we found earlier.

Example. The cost in dollars of producing $x$ stereos is given by $C(x)=70x+800$. The demand equation is $20p+x=18000$. (a) What level of production maximizes profit? (b) What is the price per stereo when profit is maximized? (c) What is the maximum profit?

Solution. From the demand equation, we obtain $p=-0.05x+900$. Recall that the profit function $P(x)$ is given by
\begin{align*} P(x)&=R(x)-C(x)\\ &=xp-C(x)\\ &=-0.05x^2+900x-(70x+800)\\ &=0.05x^2+830x-800 \end{align*}
Setting $P'(x)=-0.1 x+830$ equal to 0, we find the critical point $x=8300$. $P^{\prime\prime}(x)=-0.1<0$, so the profit has the maximum at $x=8300$.

(a) The level of production that maximizes profit is $x=8300$.

(b) The price per stereo at which profit is maximized is
$$p=-0.05(8300)+900=485\ \mathrm{dollars}$$

(c) The maximum profit is
$$P(8300)=-0.05(8300)^2+830(8300)-800=3,443,700\ \mathrm{dollars}$$

Here is another interesting theorem from economics.

Theorem. The profit is maximized when the marginal revenue equals the marginal cost, that is, when $R'(x)=C'(x)$.

Proof. Differentiating revenue function $P(x)=R(x)-C(x)$, we have
$$P'(x)=R'(x)-C'(x)$$
The critical point is obtained from $P'(x)=0$, i.e. when $R'(x)-C'(x)=0$ or $R'(x)=C'(x)$. That is, when the marginal revenue equals the marginal cost. This completes the proof.

The preceding example can be answered quickly using this theorem. Setting $R'(x)=C'(x)$, we have
$$-0.1x+900=70$$
or
$$x=8300$$

Example. A theater has 204 seats. The manager finds that he can fill all the seats if he charges \$4.00 per ticket. For each ten cents that he raises the ticket price he will sell three fewer seats. What ticket price should he charge to maximize the ticket revenue?

Solution. When the manager increases $n$ cents per ticket price, the number $x(n)$ of tickets sold is $x(n)=204-3n$ and the price $p(n)$ per ticket is $p(n)=4+0.1n$. Then the total revenue is given by
\begin{align*} R(n)&=x(n)p(n)\\ &=(204-3n)(4+0.1n)\\ &=-0.3n^2+8.4n+816 \end{align*}
Setting $R'(n)=-0.6n+8.4$ equal to 0, we find $n=14$. Since $R^{\prime\prime}(n)=-0.6<0$, the ticket revenue is maximized when $n=14$, i.e. when the ticket price is \$4.00+\$1.40=\$5.40.

Alternatively, one can easily find the demand equation which is linear in this case. The equation of line through two points $(204,4)$ and $(201,4.1)$ is given by
$$p=-\frac{1}{30}x+\frac{54}{5}$$
and so we obtain the revenue function
$$R(x)=-\frac{1}{30}x^2+\frac{54}{5}x$$
Setting $R'(x)=-\frac{x}{!5}+\frac{54}{5}$ equal to 0, we find $x=162$ and plugging this into the demand equation for $x$ gives $p=5.40$.

Let us consider a demand equation given as $x=D(p)$. If one were to consider $\frac{dx}{dp}$, the rate of change of demand with respect to price, often it would be convenient to have it as a dimensionless quantity, i.e. one that does not depend on particular units. For that we define a new quantity by dividing $\frac{dx}{dp}$ by $\frac{x}{p}$. The resulting ratio is called the elasticity of demand and is denoted by $\epsilon_D$.

Definition. The elasticity of demand $\epsilon_D$ is defined by
$$\epsilon_D=\frac{\frac{dx}{dp}}{\frac{x}{p}}=\frac{p}{x}\frac{dx}{dp}$$

In economics, demand decreases as price increases, so demand function is a decreasing function, i.e. $\frac{dx}{dp}<0$. Since both $x$ and $p$ are positive, the elasticity $\epsilon_D$ is always negative. The demand is said to be elastic if $|\epsilon_D|>1$, inelastic if $|\epsilon_D|<1$, and unitary if $|\epsilon_D|=1$.

Definition. The relative change of a function whose equation is $p=f(a)$ as $q$ changes from $q_1$ to $q_2$ is
$$\frac{f(q_2)-f(q_1)}{f(q_1)}$$
The percentage change is defined as
$$100\times\frac{f(q_2)-f(q_1)}{f(q_1)}$$

Example. (a) If the demand equation is $x=100-3p$ , find the elasticity of demand when $p=1$.

(b) Show that the elasticity equals the ratio of the relative change in demand to the relative change in price when $p$ changes from 1 to 2.

Solution. (a) $x=100-3p$, $\frac{dx}{dp}=-3$, and when $p=1$, $x=97$, so
$$\epsilon_D=\frac{p}{x}\frac{dx}{dp}=-\frac{3}{97}$$
Since $|\epsilon_D|<1$, the demand is inelastic.

(b) As $p$ changes from 1 to 2, the relative change in price is $\frac{2-1}{1}=1$. When $p$ changes from 1 to 2, $x$ changes from 97 to 94, so the relative change in demand is $-\frac{3}{97}$. Hence, the ration of the relative change in demand to the relative change in price is
$$\frac{-\frac{3}{97}}{1}=-\frac{3}{97}=\epsilon_D$$

Example. Given the demand equation $x=\sqrt{100-2p}$, find the elasticity of demand when $p=18$. Is the demand elastic or inelastic at $p=18$?

Solution. The elasticity $\epsilon_D$ at $p=18$ is
\begin{align*} \epsilon_D&=\frac{p}{x}\frac{dx}{dp}\\ &=\frac{18}{\sqrt{100-2(18)}}\left(-\frac{1}{\sqrt{100-2p}}\right)_{p=18}\\ &=\frac{18}{8}\left(-\frac{1}{8}\right)\\ &=-\frac{9}{32} \end{align*}
Since $|\epsilon_D|=\frac{9}{32}<1$, the demand at $p=18$ is inelastic.

Example. Show that when the revenue is maximized, $|\epsilon_D|=1$.

Solution. The total revenue is $R=xp$, so
\begin{align*} \frac{dR}{dp}&=\frac{dx}{dp}p+x\\ &=x\left(1+\frac{p}{x}\frac{dx}{dp}\right)\\ &=x(1+\epsilon_D) \end{align*}
Since $x>0$, $\frac{dR}{dp}=0$ if and only if $\epsilon_D=-1$. If $\epsilon_D<-1$, then $|\epsilon_D|>1$ i.e. the demand is elastic, and $\frac{dR}{dp}<0$ (the revenue is decreasing). If $-1<\epsilon_D<0$, then $|\epsilon_D|<1$ i.e. the demand is inelastic, and $\frac{dR}{dp}>0$ (the revenue is increasing). So, the revenue is maximized when $|\epsilon_D|=1$.

When demand is a function of price, the elasticity can be written as
$$\epsilon_D=\frac{\frac{p}{x}}{\frac{dp}{dx}}$$

Born Rule

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Let $A$ be a Hermitian (self-adjoint) operator. Due to Max Born, who proposed the statistical interpretation of quantum mechanics, the probability of measuring an eigenvalue $\lambda_i$ of $A$ in a state $\psi$ is $\langle\psi|P_i|\psi\rangle$, where $P_i$ is the projection onto the eigenspace of $A$ corresponding to $\lambda_i$ i.e. $P_i$ is a linear map $P_i: V_{\lambda_i}\longrightarrow V_{\lambda_i}$ such that $P^2=I$. If we assume no degeneracy of $\lambda_i$, then the eigenspace of $A$ corresponding to $\lambda_i$ is one-dimensional. In this case, $P_i=|\lambda_i\rangle\langle\lambda_i|$, where the $|\lambda_i\rangle$ are orthonormal. Now,
\begin{align*} \langle\psi|P_i|\psi\rangle&=\langle\psi|\lambda_i\rangle\langle\lambda_i|\psi\rangle\\ &=\overline{\langle\lambda_i|\psi\rangle}\langle\lambda_i|\psi\rangle\\ &=|\langle\lambda_i|\psi\rangle|^2 \end{align*}
The complex number $\langle\lambda_i|\psi\rangle$ is called the probability amplitude, so the probability is the squared magnitude of the amplitude. This is called the Born rule. (It is named after Max Born.) Suppose $|\psi\rangle$ is a unit vector. Then since $\sum_i|\langle\lambda_i|\psi\rangle|^2=1$, we have \begin{equation}\label{eq:complete}\sum_iP_i=\sum_i|\lambda_i\rangle\langle\lambda_i|=I\end{equation} \eqref{eq:complete} is called the completeness of the $|\lambda_i\rangle$.

One may consider $A$ as a random variable with its eigenvalues as the values of random variable. (See Remark below.) Since for each $j$, $A|\lambda_j\rangle=\lambda_j|\lambda_j\rangle$,
$$\langle\lambda_i|A|\lambda_j\rangle=\lambda_j\langle\lambda_i|\lambda_j\rangle=\lambda_j\delta_{ij}$$
The expected value $\langle A\rangle$ of the self-adjoint operator $A$ in the state $|\psi\rangle$ is naturally defined as a weighted average
\begin{align*} \langle A\rangle&=\sum_i\lambda_i|\langle\lambda_i|\psi\rangle|^2\\ &=\sum_i\langle\lambda_i|\psi\rangle\langle\psi|\lambda_i|\lambda_i\rangle\\
&=\sum_i\langle\lambda_i|\psi\rangle\langle\psi|A|\lambda_i\rangle\ (A|\lambda_i\rangle=\lambda_i|\lambda_i\rangle)\\
&=\sum_i\langle\psi|A|\lambda_i\rangle\langle\lambda_i|\psi\rangle\\&=\langle\psi|A|\sum_i|\lambda_i\rangle\langle\lambda_i|\psi\rangle\\&=\langle\psi|A|I|\psi\rangle\ (\mbox{by the completeness of the}\ |\lambda_i\rangle)\\&=\langle\psi|A|\psi\rangle \end{align*}
Hence, we have
$$\langle A\rangle=\langle\psi|A|\psi\rangle$$

Remark. Let $\Omega=\bigcup_iV_{\lambda_i}$, i.e. $\Omega$ is the set of all eigenvectors of $A$. Let $\mathcal{U}$ consist of $\emptyset$, $\Omega$ and unions of subsequences of ${V_{\lambda_i}}$. Then $\mathcal{U}$ is a $\sigma$-algebra of subsets of $\Omega$. Define $X:\Omega\longrightarrow\mathbb{R}$ by
$$X|\lambda_i\rangle=\langle\lambda_i|A|\lambda_i\rangle=\lambda_i$$
Then $X$ is a random variable.

Quantum Degeneracy

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In quantum mechanics, an energy level is said to be degenerate if the energy corresponds to two or more states of a quantum system. Also, two or more states of a quantum system are said to be degenerate if they give the same energy upon measurement. The set of all degenerate states of a quantum system that correspond a particular energy $E$ forms a (Hilbert) subspace, called the eigenspace of $E$. To see this, let $H$ be a Hamiltonian and $|\psi_1\rangle$, $|\psi_2\rangle$ two linearly independent eigenstates corresponding to the same eigenvalue (energy) $E$. Then
\begin{align*} H|\psi_1\rangle&=E|\psi_1\rangle\\ H|\psi_2\rangle&=E|\psi_2\rangle \end{align*}
Let $|\psi\rangle=c_1|\psi_1\rangle+c_2|\psi_2\rangle$, a linear combination (superposition) of $|\psi_1\rangle$ and $|\psi_2\rangle$. Then
\begin{align*} H|\psi\rangle&=H(c_1|\psi_1\rangle+c_2|\psi_2\rangle)\\ &=c_1H|\psi_1\rangle+c_2H|\psi_1\rangle\\&=c_1E|\psi_1\rangle+c_2E|\psi_2\rangle\\ &=E(c_1|\psi_1\rangle+c_2|\psi_2)\\ &=E|\psi\rangle \end{align*}
The dimension of the eigenspace corresponding to an eigenvalue (energy) $E$ is called the degree of degeneracy of $E$.

Brachistochrone Curve

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Brachistochrone Problem: Find the shape of the curve down which a bead sliding from rest and accelerated by gravity will slip from on point to another in the least time. Here, we do not consider friction.

Due to conservation of energy, we have
\begin{equation}
\label{eq:energy}
\frac{1}{2}mv^2=mgy
\end{equation}
Solving \eqref{eq:energy} for the speed $v$, we obtain
$$v=\sqrt{2gy}$$
The time $T_{PQ}$ for the bead to travel from a point $P$ to $Q$ is
\begin{align*}t_{PQ}&=\int_P^Q\frac{ds}{v}\\&=\int_P^Q\frac{\sqrt{dx^2+dy^2}}{\sqrt{2gy}}\\&=\int_P^Q\sqrt{\frac{1+y_x^2}{2gy}} \end{align*}
Let $f(y_x,y)=\sqrt{\frac{1+y_x^2}{2gy}}$. The path from $P$ to $Q$ which minimizes $t_{PQ}$ can be found by solving the Euler-Lagrange equation ([1])
\begin{equation}
\label{eq:E-L}
\frac{\partial f}{\partial y}-\frac{d}{Dx}\frac{\partial f}{\partial y_x}=0
\end{equation}
It can be easily shown that \eqref{eq:E-L} is equivalent to
\begin{equation}
\label{eq:E-L2}
\frac{\partial f}{\partial x}-\frac{d}{dx}\left(f-y_x\frac{\partial f}{\partial y_x}\right)=0
\end{equation}
Since $f$ does not explicitly depend on $x$, $\frac{\partial f}{\partial x}=0$, so from \eqref{eq:E-L2}, this leads to
$$f-y_x\frac{\partial f}{\partial y_x}=C$$
for some constant $C$. On the other hand, we have
$$f-y_x\frac{\partial f}{\partial y_x}=\frac{1}{\sqrt{2gy(1+y_x^2}}$$
Hence, we arrive at the differential equation
$$\frac{dy}{dx}=\sqrt{\frac{k^2-y}{y}}$$
where $k^2=\frac{1}{2gC^2}$. The differential equation is separable and it can be written as
\begin{equation}
\label{eq:de}
\sqrt{\frac{y}{k^2-y}}dy=dx
\end{equation}
Let $\sqrt{\frac{y}{k^2-y}}=\tan\frac{\theta}{2}$. Then
$$y=k^2\sin^2\frac{\theta}{2}=\frac{1}{2}k^2(1-\cos\theta)$$
The equation \eqref{eq:de} becomes
$$k^2\sin^2\frac{\theta}{2}d\theta=dx$$
Integrating both sides with the half-angle formula $\sin^2\frac{\theta}{2}=\frac{1-\cos\theta}{2}$, we obtain
$$x=\frac{1}{2}k^2(\theta-\sin\theta)+C_1$$
for some constant $C_1$. With the condition $P(0,0)$, i.e. $x=y=0$ when $\theta=0$, we have $C_1=0$ and so,
$$x=\frac{1}{2}k^2(\theta-\sin\theta)$$
Therefore, the curve on which the bead is sliding down in the shortest time is a cycloid given by the parametric equations
\begin{align*}x&=\frac{1}{2}k^2(\theta-\sin\theta)\\y&=\frac{1}{2}k^2(1-\cos\theta)\end{align*}
In geometry, a cycloid is the curve traced by a point on a circle as it rolls along a straight line without slipping.

Figure 1. Cycloid with $k=1$ and $0\leq\theta\leq 2\pi$.

References:

  1. George Arfken, Mathematical Methods for Physicists, Third Edition, Academic Press, 1985

Bell’s Theorem

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Suppose that Boris and Natasha are in different locations far away from each other. Their mutual friend Victor prepares a pair of particles and send one each to Boris and Natasha. Boris chooses to perform one of two possible measurements, say $A_0$ and $A_1$, associated with physical properties $P_{A_0}$ and $P_{A_1}$ of the particle he received. Each $A_0$ and $A_1$ has $+1$ or $-1$ for the outcomes of measurement. When Natasha receives one of the particles, she as well chooses to perform one of two possible measurements $B_0$, $B_1$, each of which has outcome $+1$ or $-1$. Let us consider the following quantity of the measurements $A_0$, $A_1$, $B_0$, $B_1$:
$$A_0B_0+A_0B_1+A_1B_0-A_1B_1=(A_0+A_1)B_0+(A_0-A_1)B_1$$
Since $A_0=\pm 1$ and $A_1=\pm 1$, either one of $A_0+A_1$ and $A_0-A_1$ is zero and the other is $\pm 2$. If the experiment is repeated over many trials with Victor preparing new pairs of particles, the expected value of all the outcomes satisfies the inequality
\begin{equation}
\label{eq:bell}
\langle A_0B_0+A_0B_1+A_1B_0-A_1B_1\rangle\leq 2
\end{equation}
Proof of \eqref{eq:bell}:
\begin{align*}\langle A_0B_0+A_0B_1+A_1B_0-A_1B_1\rangle&=\sum_{A_0,A_1,B_0,B_1}p(A_0,A_1,B_0,B_1)(A_0B_0+A_0B_1+A_1B_0-A_1B_1)\\&\leq \sum_{A_0,A_1,B_0,B_1}2p(A_0,A_1,B_0,B_1)\\&=2 \end{align*}
The inequality \eqref{eq:bell} is a variant of the Bell inequality called the CHSH inequality. CHSH stands for John Clauser, Michael Horne, Abner Simony and Richard Holt. The derivation of the CHSH inequality \eqref{eq:bell} depends on two assumptions:

  1. The physical properties $P_{A_0}$, $P_{A_1}$, $P_{B_0}$, $P_{B_1}$ have definite values $A_0$, $A_1$, $B_0$, $B_1$ which exist independently of observation or measurement. This is called the assumption of realism.
  2. Boris performing his measurement does not influence the result of Natasha’s measurement. This is called the assumption of locality.

These two assumptions together are known as the assumptions of local realism. Surprisingly this intuitively innocuous inequality can be violated in quantum mechanics. Here is an example. Let $|0\rangle=\begin{pmatrix}
1\\
0
\end{pmatrix}$ and $|1\rangle=\begin{pmatrix}
0\\
1
\end{pmatrix}$. Then $|0\rangle$ and $|1\rangle$ are the eigenstates of
$$\sigma_z=\begin{pmatrix}
1 & 0\\
0 & -1
\end{pmatrix}$$
Victor prepares a quantum system of two qubits in the state
\begin{align*}|\psi\rangle&=\frac{|0\rangle\otimes |1\rangle-|1\rangle\otimes |0\rangle}{\sqrt{2}}\\ &=\frac{|01\rangle – |10\rangle}{\sqrt{2}} \end{align*}
He passes the first qubit to Boris, and the second qubit to Natasha. Boris measures either of the observables
$$A_0=\sigma_z,\ A_1=\sigma_x=\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}$$
and Natasha measures either of the observables
$$B_0=-\frac{\sigma_x+\sigma_z}{\sqrt{2}},\ B_1=\frac{\sigma_x-\sigma_z}{\sqrt{2}}$$
Since the system is in the state $|\psi\rangle$, the average value of $A_0\otimes B_0$ is
\begin{align*}\langle A_0\otimes B_0\rangle&=\langle\psi|A_0\otimes B_0|\psi\rangle\\ &=\frac{1}{\sqrt{2}} \end{align*}
Similarly, the average values of the other observables ar given by
\begin{align*}\langle A_0\otimes B_1\rangle&=\frac{1}{\sqrt{2}},\ \langle A_1\otimes B_0\rangle=\frac{1}{\sqrt{2}},\ \mbox{and}\\ \langle A_1\otimes B_1\rangle&=-\frac{1}{\sqrt{2}} \end{align*}
Since the expected value is linear, we have
\begin{equation}
\begin{aligned}
\langle A_0B_0+A_0B_1+A_1B_0-A_1B_1\rangle&=\langle A_0B_0\rangle+\langle A_0B_1\rangle+\langle A_1B_0\rangle-\langle A_1B_1\rangle\\&=2\sqrt{2}\end{aligned}\label{eq:bell2}
\end{equation}
This means that the Bell inequality \eqref{eq:bell} is violated. Physicists have confirmed the prediction in \eqref{eq:bell2} by experiments using photons. It turns out that the Mother Nature does not obey the Bell inequality. What this means is that one or both of the two assumptions for the derivation of the Bell inequality \eqref{eq:bell} must be incorrect. There is no consensus among physicists which of the two assumptions needs to be dropped. An important lesson we learn from the Bell inequality is that the Mother Nature (Quantum Mechanics) defies our intuitive common sense. This also begs a troubling question. If we cannot rely on our intuition to understand how the universe works, what else can we rely on? One thing is certain. The world is not locally realistic.

References:

  1. Michael A. Nielsen and Isaac L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 2004