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

$$\epsilon_D=\frac{\frac{p}{x}}{\frac{dp}{dx}}$$