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Section 3.2 Applications to Business and Economics

Just like the process of differentiation is a useful tool in many business and economics applications such as problems related to elasticity of demand and optimization, so is the process of antidifferentiation or integration. In this section, we revisit two important business and economic models, namely concerning law of supply and demand in a free-market environment and continuous money flow, and introduce integration as a method for solving problems of this nature.

Subsection 3.2.1 Surplus in Consumption and Production

Recall that in a free market, the consumer demand for a particular commodity is dependent on the commodity's unit price, which is captured graphically by the demand curve. As expected, the quantity demanded of a commodity increases as the commodity's unit price decreases, and vice versa. Similarly, the unit price of a commodity is dependent on the commodity's availability in the market, which is articulated graphically by the supply curve. Typically, an increase or decrease in the commodity's unit price induces the producer to respectively increase or decrease the supply of the commodity. Below we see the supply and demand curves of a certain item produced and sold:

Figure 3.6. Example of a supply curve (in blue) and a demand curve (in red). The point of intersection \((x_e, p_e)\) corresponds to market equilibrium.

Interactive Demonstration. Use the control points below to change the producer and supplier surpluses (the equilibrium point is fixed).

In a competitive market, the price of a commodity will eventually settle at the market equilibrium, which occurs when the supply of the commodity will be equal to its demand as indicated with the point \((x_e, p_e)\) in Figure 3.6. Let us look at a certain commodity, say, dairy to discuss the economic significance of the market equilibrium. As long as \(p \lt p_e\text{,}\) then the demand for dairy exceeds its supply (see Figure 3.6), which pushes up the price until it reaches the equilibrium price \(p_e\text{,}\) which in turn signifies that the quantity supplied is equal to the quantity demanded, namely \(x_e\text{.}\) On the other hand, if \(p > p_e\text{,}\) then the supply of dairy exceeds demand (see Figure 3.6), which brings the price down. In an ideal free market, buying and selling at the equilibrium price should benefit both consumers and producers. In this section, we will compute the surplus, which tells us exactly how much the consumers save and the producers gain by buying and selling respectively at the equilibrium price rather than at a higher price.

We begin by computing exactly how much consumers spent when they buy at the equilibrium price \(p_e\text{:}\)

\begin{equation*} \begin{split} \amp \text{ total amount spent at equilibrium price } \\ \amp = (\text{ number of units bought at equilibrium price } )\cdot(\text{ unit price } )\\ \amp =x_e \cdot p_e \end{split} \end{equation*}

The quantity \(x_e \cdot p_e\) is the rectangular area shown in Figure 3.(a).

Now we compute the total amount that would be spent if every consumer paid the maximum price that each is willing to pay. Given a demand function \(D\text{,}\) partition the interval \([0, x_e]\) into \(n\) subintervals of equal width \(\Delta x = \frac{x_e}{n}\) with endpoints \(x_i = \frac{ix_e}{n}\text{,}\) \(i=0,1,\dots, n\) as shown below:

Now, let us analyze this partition subinterval by subinterval. Suppose that only \(x_1\) units had been available, then the maximum unit price could have been set at \(D(x_1)\) dollars and a total of \(x_1\) units would be sold, but at this price no further units would have been sold. Then the total expenditure in dollars is given by

\begin{equation*} (\text{ price per unit } )\cdot(\text{ number of units } ) = D(x_1) \cdot\Delta x\text{.} \end{equation*}

Now suppose that more units become available by producing \(x_2\) units of our commodity. If the maximum price is set at \(D(x_2)\) dollars, then the remaining \(x_2 - x_1 = \Delta x\) units can be sold at a cost of \(D(x_2) \cdot \Delta x\) dollars. If we continue with this process of price discrimination, then the total amount spent at maximum price is approximately equal to

\begin{equation*} D(x_1)\cdot\Delta x + D(x_2)\cdot\Delta x + \dots + D(x_n)\cdot\Delta x\text{.} \end{equation*}

Note:

  1. On each of the subintervals \([0,x_1]\text{,}\) \([x_1,x_2]\text{,}\) \(\dots\text{,}\) \([x_{n-1},x_e]\text{,}\) the buyers paid as much for each unit as it was worth to them.

  2. We recognize that the last sum is a Riemann sum, which yields \(\ds\int_0^{x_e} D(x)\,dx\) as \(n \to \infty\) or alternatively as \(\Delta x \to 0\text{.}\)

Of course, we simply have calculated the area under the demand curve on the interval \([0, x_e]\text{,}\) which is shown in Figure 3.(b).

(a)
(b)
(c)
Figure 3.7.

The difference between the total amount spent at the maximum price (Figure 3.(b)) and the consumer expenditure at the equilibrium price (Figure 3.(a)) is the total amount that consumers save by buying at the equilibrium price (Figure 3.(c)). This saving is called the consumer surplus for this product. We summarize our result.

In a similar manner, we can determine how much producers gain when they sell at the equilibrium price \(p_e\text{:}\)

\begin{equation*} \begin{split} \amp \text{ total amount gained at equilibrium price } \\ \amp = (\text{ number of units sold at equilibrium price } ) \cdot (\text{ unit price } )\\ \amp = x_e \cdot p_e \end{split} \end{equation*}

Now we compute the total amount that would be gained if every producer sold at the minimum amount they are willing to accept for the product. Given a supply function \(S\text{,}\) partition the interval \([0, x_e]\) into \(n\) subintervals of equal width \(\Delta x = \frac{x_e}{n}\) with endpoints \(x_i = \frac{ix_e}{n}\text{,}\) \(i=0,1,\dots, n\) as shown below:

Similarly to the subinterval by subinterval analysis for the demand curve, an analysis of the partition of the supply curve shows that the total revenue on every subinterval in dollars by selling at the minimum price is given by

\begin{equation*} (\text{ price per unit) } \cdot (\text{ number of units } ) = S(x_i) \cdot \Delta x \end{equation*}

for \(i=1,2,\dots,n\text{.}\) Continuing with the process of price discrimination, the total amount gained at minimum price is approximately equal to

\begin{equation*} S(x_1)\cdot\Delta x + S(x_2)\cdot\Delta x + \dots + S(x_n)\cdot \Delta x\text{.} \end{equation*}

Note:

  1. On each of the subintervals \([0,x_1]\text{,}\) \([x_1,x_2], \dots [x_{n-1},x_e]\text{,}\) the producers sold at the lowest price that they are willing to set.

  2. This sum is again a Riemann sum, which yields \(\ds \int_0^{x_e} S(x)\, dx\) as \(n\to\infty\) or alternatively as \(\Delta x \to 0\text{.}\)

Figure 3.(b) shows the area under the supply curve on the interval \([0, x_e]\) that our computation has yielded.

(a)
(b)
(c)
Figure 3.8. Producer revenue on \([0, x_e]\) at equilibrium price, in total, and for surplus.

The difference between the producer revenue at the equilibrium price (Figure 3.(a)) and the total amount achieved at the minimum price (Figure 3.(b)) is the total amount that producers gain by selling at the equilibrium price (Figure 3.(c)). This income is called the producer surplus for this product. We summarize our result.

Note: In general, the consumer surplus and the producer surplus are not equal.
Example 3.12. Consumer Surplus.

The demand for a product, in dollars, is

\begin{equation*} p = D(x) = 1000 - 0.5x-0.0002x^2\text{.} \end{equation*}

Find the consumer surplus when the sales level is \(200\text{.}\)

Solution

When the number of units sold is \(x_e = 200\text{,}\) the corresponding price is

\begin{equation*} p_e = 1000 - 0.5(200) - 0.0002(200^2) = 892\text{.} \end{equation*}

Therefore, the consumer surplus is

\begin{equation*} \begin{split} \int_0^{200} \left[D(x) - p_e\right]\,dx \amp = \int_0^{200} \left(1000 - 0.5x-0.0002x^2 - 892\right)\, dx\\ \amp = \int_0^{200} \left(108 - 0.5x - 0.0002x^2\right)\, dx\\ \amp = \left[108x-0.25x^2 - \frac{0.0002}{3} x^3 \right]_0^{200}\\ \amp = 108(200)-0.25(200)^2 - \frac{0.0002}{3} (200)^3\\ \amp = \$ 11,066.70 \end{split} \end{equation*}
Example 3.13. Consumer and Producer Surplus.

The price, in dollars per kg, of flour at a certain grocer is given by

\begin{equation*} D(x)= 75 - e^{0.2x}, \text{ for } x \in [0,21] \end{equation*}

corresponding to a demand of \(x\) kg. When the supply is \(x\) kg, the wholesale price is given by

\begin{equation*} S(x)=2e^{0.2x}-1\text{.} \end{equation*}

Determine the consumer surplus and the producer surplus.

Solution

We set \(D(x)=S(x)\) to find the equilibrium quantity and price:

\begin{equation*} \begin{split} 75 - e^{0.2x} \amp =2e^{0.2x}-1\\ e^{0.2x} \amp = \frac{76}{3} \\ x \amp = \frac{\ln\left(\frac{76}{3}\right)}{0.2} \approx 16.16 \ \text{ kg } . \end{split} \end{equation*}

Therefore, the equilibrium point is

\begin{equation*} (x_e,p_e) = (16.16, 75-e^{0.2(16.16)}) \approx (16.16, 49.67)\text{.} \end{equation*}

We now calculate the consumer surplus as

\begin{equation*} \begin{split} \int_0^{16.16} \left[(75-e^{0.2x})-49.67\right]\,dx \amp = \int_0^{16.16} \left(25.33-e^{0.2x}\right)\,dx\\ \amp = \left[25.33x - \frac{1}{0.2}e^{0.2x}\right]_0^{16.16} \\ \amp \approx \$ 287.68. \end{split} \end{equation*}

Similarly, we calculate the producer surplus as

\begin{equation*} \begin{split} \int_0^{16.16} \left[49.67 - \left(2e^{0.2x}-1\right)\right]\,dx \amp = \int_0^{16.16} \left(50.67 - 2e^{0.2x}\right)\,dx \\ \amp = \left[50.67x - \frac{2}{0.2}e^{0.2x}\right]_0^{16.16} \\ \amp \approx \$ 575.53. \end{split} \end{equation*}

Thus, the consumer surplus is approximately $287.68 and the producer surplus is approximately $575.53. These quantities are illustrated in the figure below.

Subsection 3.2.2 Continuous Money Flow

Another application of integrals is for the computation involved in continuous money flow problems. These problems may be payments to repay a debt or deposits to establish a retirement plan that are made at regular monthly payments over a certain time period. If the time between deposits or payments are made in relatively short intervals compared to the entire time period, then the money can be thought of as flowing continuously, hence the name continuous money flow. Some textbooks refer to these types of problems as “continuous income streams” or “continuous money streams”. Naturally, deposits or payments cannot be made continuously in a physical manner, since the units of legal tender are by their very nature discrete such as the cent and dollar.

A word of caution: Discussing the value of money is difficult as there are many parameters affecting it. For the purpose of this introductory section, we take a simplified point of view such as assuming that there is a single interest rate, no inflation, and no other factors influencing computation. This simplified approach will provide for an intuitive understanding of solving continuous money flow problems that can then be expanded upon. Before we develop formulas for the present and future value of such a continuous money flow, we begin by computing the total amount of money that is being accumulated making the regular deposits or payments.

(a) A uniform rate of money flow \(f(t)=m\) dollars per year.
(b) A variable rate of money flow \(f(t)\) dollars per year.
Figure 3.9.

Suppose that \(f(t)\) is the rate in dollars per unit of time for a continuous money flow. Figure 3.9 above shows a uniform rate of money flow \(f(t)=m\) dollars per year to the left and a variable rate of money flow \(f(t)\) dollars per year to the right. The total income in dollars of the former case can be readily calculated and is simply the rectangular area \(m\cdot T\) dollars. For the latter case, we readily understand that the area under the curve \(f\) from \(t=0\) to \(t=T\) is the total income, whose value in dollars is determined by the definite integral \(\ds \int_0^T f(t)\,dt\text{.}\)

Note: The above computation does not calculate the accumulated interest, just the total income.
Example 3.15. Total Income.

Suppose the income of a small coffee shop projected to grow exponentially. The rate of income when the shop first opened was at $5,000 per month, which grew to a rate of $6,000 per month by the second month. What is the expected total income from the coffee shop during its first 6 months of operation?

Solution

Let \(t\) be the time in months since the opening of the shop. The projected income is of the form

\begin{equation*} f(t) = C e^{kt}\text{.} \end{equation*}

Since \(f(0) = 5000\text{,}\) we must have \(C=5000\text{.}\) Now to determine \(k\text{,}\) we use \(f(1)=6000\text{:}\)

\begin{equation*} \begin{split} 5000 e^{k(1)} \amp = 6000 \\ k \amp = \ln \frac{6000}{5000} \\ k \amp \approx 0.18. \end{split} \end{equation*}

Thus, \(f(t) = 5000e^{0.18 t}\text{.}\) We can now determine the total income, \(I\text{:}\)

\begin{equation*} \begin{split} I \amp = \int_0^6 f(t) \, dt \\ \amp = \int_0^6 5000e^{0.18 t} \, dt\\ \amp = \frac{5000}{0.18} e^{0.18 t} \big\vert_0^6\\ \amp \approx 54018.8764. \end{split} \end{equation*}

We conclude that the coffee shop is expected to produce an income of approximately $54,018.88 in its first six months of operation.

Let us now develop the formulas for the present value and future value of a continuous money flow. We begin by summarizing the present and future value of money that earns interest compounded continuously at a rate \(r\) over time t without making continuous deposits or payments. Suppose that \(F\) is the amount of money at time \(t\) that accumulates interest compounded continuously at a rate \(r\) over time \(t\text{,}\) then the growth of the investment or loan due to accumulated interest earned over time is described by the differential equation

\begin{equation*} \frac{dF}{dt} = rF\text{.} \end{equation*}

In Chapter 5, differential equations will be discussed in more detail, but for now, we simply present the solution to this differential equation, and so the future value is described by

\begin{equation*} F(t) = Pe^{rt}\text{,} \end{equation*}

where \(P\) is the initial amount. Similarly, if we want to obtain an amount of \(F\) dollars that has accrued interest compounded continuously at the annual rate \(r\) over \(t\) years, then the amount of money that needs to be deposited today, namely the present value \(P\text{,}\) is given by

\begin{equation*} P(t) = Fe^{-rt}\text{.} \end{equation*}

Now, the present value \(P\) of a continuous money stream must be the amount of money that will accumulate to the same future value as the continuous money when deposited at the same interest rate compounded continuously for the same period of time but without the continuous stream of income. Let \(f(t)\) represent the rate of continuous money flow and \(r\) the rate of interest for \(T\) years. As usual, we partition the interval \([0, T]\) into \(n\) subintervals of equal width \(\Delta t = \frac{T}{n}\) with endpoints \(t_i = \frac{iT}{n}\text{,}\) \(i=0,1,\dots,n\) as shown below.

Using our result from above about accruing interest compounded continuously, the present value \(P_i\) of the money flow over the \(i\)-th subinterval is approximately

\begin{equation*} P_i = \left[f(t_i)\Delta t\right] e^{-rt_i}\text{,} \end{equation*}

and so the total present value is approximated by a typical Riemann sum

\begin{equation*} P\approx\sum_{i=1}^n\left[f(t_i)\Delta t\right] e^{-rt_i}\text{.} \end{equation*}

Taking the limit as \(n\to\infty\text{,}\) we obtain the following definite integral

\begin{equation*} P = \int_0^T f(t) e^{-rt}\,dt\text{.} \end{equation*}

We capture our result in the following theorem.

Before we develop the formula for the accumulated amount of a continuous money flow, we present one example of a present value problem.

Example 3.17. Present Value.

A certain manufacturer produces industrial bread-making machines, which are estimated to generate a continuous income stream each with a rate of \(28 - 4.5t\) million dollars per year in year \(t\) of operation. Each machine is in operation for about 6 years and the money is invested at the annual rate of 5% compounded continuously. Compute the fair market price of each machine.

Solution

We assume that the fair market price of each machine is the present value of the continuous money flow with

\begin{equation*} f(t) = 28-4.5t , \ r=0.05, \ \text{ and } \ T=6\text{.} \end{equation*}

Therefore, the price \(P\) is given by

\begin{equation*} P = \int_0^6 \left[28-4.5t\right]e^{-0.05t}\,dt = \int_0^6 28e^{-0.05t}\,dt -4.5\int_0^6 te^{-0.05t}\,dt\text{.} \end{equation*}

The first integral can be computed directly

\begin{equation*} \int_0^6 28e^{-0.05t}\,dt = \left[-\frac{28}{0.05}e^{-0.05t}\right]_0^6 = 560\left(1-e^{-0.03}\right)\text{.} \end{equation*}

The second integral can be solved using Integration by Parts, with

\begin{equation*} u=t, \ dv = e^{-0.05t} dt \implies du = dt, \ v = -\frac{1}{0.05}e^{-0.05t} \end{equation*}

to obtain

\begin{equation*} \begin{split} \int_0^6 te^{-0.05t}\,dt \amp = \left[-\frac{t}{0.05}e^{-0.05t}\right]_0^6 + \frac{1}{0.05} \int_0^6 e^{-0.05t}\,dt \\ \amp = - \left[\frac{0.05t+1}{0.05}e^{-0.05t}\right]_0^6 = 20-26e^{-0.3}. \end{split} \end{equation*}

Hence,

\begin{equation*} \begin{split} P \amp = \int_0^6 \left[28-4.5t\right]e^{-0.05t}\,dt \\ \amp = 560 \left(1-e^{-0.3}\right) - 4.5 \left(20-26e^{-0.3}\right) \\ \amp \approx 141.82, \end{split} \end{equation*}

and so the fair market price of each machine is about 142 million dollars.

Similarly, we can develop a formula for the future value of a continuous money flow. Again, we partition the interval \([0, T]\) into \(n\) subintervals of equal width \(\Delta t = \frac{T}{n}\) with endpoints \(t_i = \frac{iT}{n}\text{,}\) \(i=0,1,\dots,n\text{.}\) Now, the money that is deposited or paid during the \(i\)-th time interval stays in the account for the length of time \(T- t_i\text{.}\) Using our result from the beginning of this section about interest being earned compounded continuously, the future value \(F_i\) of the money flow over the \(i\)-th subinterval is approximately

\begin{equation*} F_i = \left[f(t_i)\Delta t\right]e^{r(T-t_i)}\text{,} \end{equation*}

and so the total future value is again approximated by a Riemann sum, namely

\begin{equation*} F \approx \sum_{i=1}^n \left[f(t_i)\Delta t\right]e^{r(T-t_i)}\text{.} \end{equation*}

Taking the limit as \(n\to\infty\text{,}\) the following definite integral is obtained

\begin{equation*} F = \int_0^T f(t)e^{r(T-t)}\,dt\text{.} \end{equation*}

This leads to the following theorem.

Example 3.19. Continuous Money Flow.

Suppose money is flowing continuously at a constant rate \(f(t)\) for 10 years at 8% interest compounded continuously accumulating to $20,000. Compute \(f(t)\text{.}\)

Solution

We have that

\begin{equation*} F=20,000, \ r=0.08, \ \text{ and } T=10\text{.} \end{equation*}

We can use the formula for the future value to solve for the rate \(f(t)\) of the continuous money flow:

\begin{equation*} 20,000 = \int_0^{10} f(t) e^{0.08(10-t)}\,dt \end{equation*}

Since the rate of money flow is constant, \(f(t)\) does not depend on \(t\text{.}\) Therefore, \(f(t)\) can be taken out of the integral as follows.

\begin{equation*} \begin{split} 20,000 \amp = f(t) \int_0^{10} e^{0.08(10-t)}\,dt \\ -0.08\cdot e^{-0.8}\cdot 20,000 \amp = f(t) \left[e^{-0.08t}\right]_0^{10} \\ -0.08\cdot e^{-0.8} \cdot 20,000 \amp = f(t) \left[e^{-0.8}-e^0\right] \\ f(t)\amp = \frac{0.08\cdot 20,000}{\left[e^{0.8}-1\right]} \approx \$1305.55 \end{split} \end{equation*}
Example 3.20. Total Income, Present and Future Values.

Suppose money is flowing continuously at a constant rate of $3500 per year for 10 years at 1.25% interest compounded continuously. Compute the following:

  1. total income;

  2. future value at \(T=10\) years;

  3. total interest; and

  4. present value.

Solution

We have that \(f(t) = 3500\text{,}\) \(r=0.0125\text{,}\) and \(T=10\text{.}\)

  1. The total income \(I\) is given by

    \begin{equation*} I = \int_0^{10} 3500 \,dt = \$35,000\text{.} \end{equation*}

  2. The future value \(F\) at \(T=10\) years is calculated to be

    \begin{equation*} F = \int_0^{10} 3500 e^{0.0125(10-t)}\,dt = -\frac{3500}{0.0125}\left[e^{0.0125(10-t)}\right]_0^{10} = \$37,281.75\text{.} \end{equation*}

  3. The total interest is the difference between the future value \(F\) at \(T=10\) years and the total income over \(10\) years:

    \begin{equation*} \$37,281.75 - \$35,000 = \$2281.57\text{.} \end{equation*}

  4. The present value \(P\) is obtained as

    \begin{equation*} \begin{split} P \amp = \int_0^{10} 3500e^{-0.0125t}\,dt \\ \amp = \left[-\frac{3500}{0.0125} e^{-0.0125}\right]_0^{10}\\ \amp = \frac{3500}{0.0125}\left(1-e^{-0.125}\right) \approx \$32,900.87. \end{split} \end{equation*}
Exercises for Section 3.2.

Suppose the supply function for a certain product is given by

\begin{equation*} S(q)=q^{5/2} + 3q^{1/2}+25\text{.} \end{equation*}

If the equilibrium quantity is \(q=10\text{,}\) determine the producer surplus.

Answer
$2290.40
Solution

If the equilibrium quantity is \(q=10\text{,}\) this means that the equilibrium price is \(p=S(10) \approx \$ 350.71\text{.}\) We make a sketch of this scenario below:

Therefore, the green shaded area represents the producer surplus, \(P\text{.}\) We compute:

\begin{equation*} \begin{split} P \amp = \int_0^{10} \left[S(10) - S(q)\right]\,dq\\ \amp = 10 \cdot S(10) - \int_0^{10} (q^{5/2}+3q^{1/2}+25) \,dq \\ \amp = 10 \cdot S(10) - \left[\frac{2q^{7/2}}{7} + 2q^{3/2} + 25 q\right]_0^{10}\\ \amp \frac{5070\sqrt{10}}{7} \approx \$2290.40 \end{split} \end{equation*}

It is determined that the supply function for a certain nutritional supplement can be described by

\begin{equation*} S(x)=x^{3/2} + 2x^{1/2} + 50\text{.} \end{equation*}

Assuming that the supply and demand are in equilibrium when \(x=12\text{,}\) determine the producer surplus.

Answer
$327.01
Solution

If the equilibrium quantity is \(x=12\text{,}\) this means that the equilibrium price is \(p=S(12) \approx \$98.50\text{.}\) We make a sketch of this scenario below:

Hence, the green shaded area denotes the producer surplus, \(P\text{.}\) We compute:

\begin{equation*} \begin{split} P \amp = \int_0^{12} \left[S(12) - S(x)\right]\,dx \\ \amp = 12\cdot S(12) - \int_0^{12} (x^{3/2}+2x^{1/2}+50) \,dx\\ \amp = \frac{944\sqrt{3}}{5} \approx \$327.01 \end{split} \end{equation*}

Determine the consumer surplus corresponding to the demand function

\begin{equation*} D(q) = \frac{100}{(q+2)^2}\text{,} \end{equation*}

if the equilibrium quantity is \(q=2\text{.}\)

Answer
$12.5
Solution

If the equilibrium quantity is \(q=2\text{,}\) this means that the equilibrium price is \(p=D(2)=\$6.25\text{.}\) We make a sketch of this scenario below:

where the red shaded area represents the consumer surplus, \(C\text{.}\) We therefore compute:

\begin{equation*} \begin{split} C\amp =\int_0^2 \left[D(q) - D(2)\right]\,dq \\ \amp = \int_0^2 \frac{100}{(q+2)^2}\,dq - \frac{50}{4}\\ \amp = -\frac{100}{q+2}\bigg\vert_0^2 - \frac{50}{4}\\ \amp = 25 - \frac{50}{4} \approx \$12.5 \end{split} \end{equation*}

Suppose the demand function for organic tomatoes can be described by

\begin{equation*} D(x) = \frac{20,000}{(2x+5)^3}\text{.} \end{equation*}

Determine the consumer surplus if the supply and demand are in equilibrium at \(x=4\text{.}\)

Answer
$90.69
Solution

If the equilibrium quantity is \(x=4\text{,}\) this means that the equilibrium price is \(p=D(4) \approx \$0.91\text{.}\) We make a sketch of this scenario below:

where the red shaded area represents the consumer surplus, \(C\text{.}\) We therefore compute:

\begin{equation*} \begin{split} C\amp =\int_0^4 \left[D(x) - D(4)\right]\,dx \\ \amp = \int_0^4 \frac{20,000}{(4x+5)^3}\,dq - 4 D(4)\\ \amp \approx \$90.69 \end{split} \end{equation*}

The supply function for steel-cut oats is found to be

\begin{equation*} S(q) = q(2q+25)\text{,} \end{equation*}

dollars and the demand function is found to be

\begin{equation*} D(q) = 1000 - 10 q - q^2\text{,} \end{equation*}

dollars, where \(q\) is measured in kgs.

  1. Determine the equilibrium point.

    Answer
    Solution

    We set \(S(q) = D(q)\) to find the equilibrium point:

    \begin{equation*} \begin{gathered} q(2q+25) = 1000-10q-q^2\\ 2q^2 + 25q = 1000-10q-q^2\\ 3q^2 +35q - 1000 = 0 \\ q = -25, q= \frac{40}{3} \end{gathered} \end{equation*}

    We reject the negative solution. Therefore, the equilibrium point is when \(q \approx 13.33\) kgs, with a corresponding price of about $688.89.

  2. What is the consumer surplus?

    Answer
    Approx. $2469.14
    Solution

    The consumer surplus is given by

    \begin{equation*} \int_0^{40/3} \left[\left(1000-10q-q^2\right) - D(40/3)\right]\,dq \approx \$2469.14\text{.} \end{equation*}

  3. What is the producer surplus?

    Answer
    Approx. $5382.72
    Solution

    The producer surplus is therefore

    \begin{equation*} \int_0^{40/3} \left[ S(40/3) - q(2q+25)\right]\,dq \approx \$5382.72\text{.} \end{equation*}

  4. Display your results on a graph of the supply and demand curves.

    Solution

    We graph the information calculated in parts (a)-(c) below:

It is determined that the supply function for a certain product \(x\) (measured in units of thousands) can be described by

\begin{equation*} S(x)=\left(2x+\frac{1}{4}\right)^2\text{,} \end{equation*}

with a corresponding demand function

\begin{equation*} D(x) = \frac{200}{15x+1}\text{.} \end{equation*}

  1. Determine the equilibrium point.

    Answer
    Approx. \((1.39,9.16)\)
    Solution

    We find the equilibrium point by setting the supply equation equal to the demand equation:

    \begin{equation*} \begin{gathered} S(x) = D(x)\\ \left(2x+\frac{1}{4}\right)^2 = \frac{200}{15x+1}\\ 4x^2+x+\frac{1}{16} = \frac{200}{15x+1}\\ 60x^3+19x^2+\frac{31x}{16}+\frac{1}{16} = 200, \end{gathered} \end{equation*}

    from which we find \(x=1.3885\text{.}\) Therefore the equilibrium point is approximately

    \begin{equation*} (1.39,S(1.39)) = (1.39,D(1.39)) \approx (1.39,9.16)\text{.} \end{equation*}

  2. What is the consumer surplus?

    Answer
    Approx. $28.39
    Solution

    The consumer surplus is given by

    \begin{equation*} \int_0^{1.39} \left[\frac{200}{15x+1} - 9.16\right]\,dx = \left[\frac{200}{15}\log|15x+1|-9.16x\right]_0^{1.39} \approx \$28.39\text{.} \end{equation*}

  3. What is the producer surplus?

    Answer
    Approx. $8.10
    Solution

    The supplier surplus is calculuated as

    \begin{equation*} \int_0^{1.39} \left[9.16-\left(2x+\frac{1}{4}\right)^2\right]\,dx = \left[9.16x-\frac{4x^3}{3}-\frac{x^2}{2}-\frac{x}{16}\right]_0^{1.39} \approx \$8.10\text{.} \end{equation*}

  4. Display your results on a graph of the supply and demand curves.

    Solution

    We graph the information calculated in parts (a)-(c) below:

For each of the following rate of money flow in dollars per year, rate \(r\) of continuously compounded interest, and time period \(T\) years, compute (i) the total income; (ii) the future value at \(T\text{;}\) (iii) the total interest; and (iv) the present value.

  1. \(f(t) = 200,\ r= 2\%, \ T=5\text{.}\)

    Answer
    \(\$ 1000; \$1051.71; \$51.71; \$951.63\text{.}\)
    Solution

    The total income of the money flow is

    \begin{equation*} I = \int_0^5 200\,dt = \$ 1000\text{.} \end{equation*}

    The future value is given by

    \begin{equation*} F = \int_0^5 200 e^{0.02(5-t)}\,dt = \frac{-200}{0.02}e^{0.02(5-t)}\bigg\vert_0^5 = \$1051.71\text{.} \end{equation*}

    Therefore, the total interest is

    \begin{equation*} F-I = \$51.71\text{.} \end{equation*}

    The present value of the money flow is

    \begin{equation*} P=\int_0^5 200 e^{-0.02t}\,dt = \$951.63\text{.} \end{equation*}

  2. \(f(t) = 4000,\ r = 0.5\%, \ T=7\text{.}\)

    Answer
    \(\$28,000; \$28,495.80; \$495.80; \$27,515.70\)
    Solution

    The total income of the money flow is

    \begin{equation*} I = \int_0^7 4000 \,dt = \$28,000\text{.} \end{equation*}

    The future value is given by

    \begin{equation*} F = \int_0^7 4000e^{0.005(7-t)}\,dt = \frac{-4000}{0.005}e^{0.005(7-t)}\bigg\vert_0^7 = \$28,495.80\text{.} \end{equation*}

    Therefore, the total interest is

    \begin{equation*} F-I = \$495.80 \end{equation*}

    The present value of the money flow is

    \begin{equation*} P=\int_0^7 4000e^{-0.005t}\,dt = \$27,515.70 \end{equation*}

  3. \(f(t)=10,000, \ r=1\%, T=4\text{.}\)

    Answer
    \(\$40,000; \$40810.80; \$810.80; \$39210.60\)
    Solution

    The total income of the money flow is

    \begin{equation*} I = \int_0^4 10000 \,dt = \$40,000\text{.} \end{equation*}

    The future value is given by

    \begin{equation*} F = \int_0^4 10000e^{0.01(4-t)}\,dt = \frac{-10000}{0.01}e^{0.01(4-t)}\bigg\vert_0^4 = \$40810.80\text{.} \end{equation*}

    Therefore, the total interest is

    \begin{equation*} F-I = \$810.80\text{.} \end{equation*}

    The present value of the money flow is

    \begin{equation*} P=\int_0^4 10000e^{-0.01t}\,dt = \$39210.60\text{.} \end{equation*}

  4. \(f(t) = 300t, \ r=1\%, T=4\text{.}\)

    Answer
    \(\$2400; \$2432.32; \$32.32; \$2336.95\text{.}\)
    Solution

    The total income of the money flow is

    \begin{equation*} I = \int_0^4 300t \,dt = \$2400\text{.} \end{equation*}

    The future value is given by

    \begin{equation*} F = \int_0^4 300t e^{0.01(4-t)}\,dt = \$2432.32\text{.} \end{equation*}

    Therefore, the total interest is

    \begin{equation*} F-I = \$32.32\text{.} \end{equation*}

    The present value of the money flow is

    \begin{equation*} P=\int_0^4 300t e^{-0.01t}\,dt = \$2336.95\text{.} \end{equation*}

  5. \(f(t) = 450e^{0.5t}, \ r=1.5\%, T=15\text{.}\)

    Answer
    \(\$1,626,240; \$1,676,400; \$50,160; \$1,338,640\text{.}\)
    Solution

    The total income of the money flow is

    \begin{equation*} I = \int_0^{15} 450e^{0.5t} \,dt = \$1,626,240\text{.} \end{equation*}

    The future value is given by

    \begin{equation*} F = \int_0^{15} 450e^{0.5t} e^{0.015(15-t)}\,dt = \$1,676,400\text{.} \end{equation*}

    Therefore, the total interest is

    \begin{equation*} F-I = \$50,160\text{.} \end{equation*}

    The present value of the money flow is

    \begin{equation*} P=\int_0^{15} 450e^{0.5t} e^{-0.015t}\,dt = \$1,338,640\text{.} \end{equation*}

  6. \(f(t) = 100t+50, \ r=3\%, \ T=9\text{.}\)

    Answer
    \(\$4500; \$4957.10; \$457.10; \$3784.15\text{.}\)
    Solution

    The total income of the money flow is

    \begin{equation*} I = \int_0^9 \bigl(100t+50\bigr) \,dt = \$4500\text{.} \end{equation*}

    The future value is given by

    \begin{equation*} F = \int_0^9 \bigl(100t+50\bigr)e^{0.03(9-t)}\,dt = \$4957.10\text{.} \end{equation*}

    Therefore, the total interest is

    \begin{equation*} F-I = \$457.10\text{.} \end{equation*}

    The present value of the money flow is

    \begin{equation*} P=\int_0^9 \bigl(100t+50\bigr) e^{-0.03t}\,dt = \$3784.15\text{.} \end{equation*}

The value of a continuous money flow is found to be $5,000 after 6 years. Assume the rate of money flow \(f(t)\) is constant, and the interest is continuously compounded at a rate of 1%. Compute \(f(t)\text{.}\)

Answer
\(f(t) \approx \$808.58\)
Solution

We are given that the value of a continuous money flow is $5000 after a period of 6 years with an interest rate of \(0.01\text{.}\) Using the Future Value formula, this means that we must have

\begin{equation*} 5000 = \int_0^6 f(t) e^{0.01(6-t}\,dt\text{.} \end{equation*}

Now if we assume that \(f(t)\) is a constant function \(C\text{,}\) we have

\begin{equation*} \begin{split} 5000 \amp = \int_0^6 C e^{0.01(6-t}\,dt \\ 5000 \amp = C \int_0^6 e^{0.01(6-t}\,dt\\ 5000 \amp = C \left[ \frac{-1}{0.01}e^{0.01(6-t)}\right]_{0}^{6}\\ 5000 \amp = C \left(6.18365\right)\\ C \amp \approx \$808.58. \end{split} \end{equation*}

The value of a continuous money flow is found to be $10,000 after 5 years, and $14,000 after 7 years. The rate of money flow \(f(t)\) is linear, and the interest is compounded continuously at a rate of 1.5%. Compute \(f(t)\text{.}\)

Answer
\(f(t) \approx -30 t + 2000\)
Solution

Since the continuous money flow is a linear function of time, let \(f(t) = at+b\text{.}\) Then, using the Future Value formula, we have that

\begin{equation*} \begin{split} 10,000 \amp = \int_0^5 f(t) e^{0.015(5-t)}\,dt = \int_0^5 (at+b)e^{0.015(5-t)}\,dt, \text{ and } \\[1ex] 14,000 \amp = \int_0^7 f(t) e^{0.015(7-t)}\,dt = \int_0^7 (at+b)e^{0.015(7-t)}\,dt. \end{split} \end{equation*}

First, consider the integral

\begin{equation*} \int_0^T te^{0.015(T-t)}\,dt\text{.} \end{equation*}

We solve using integration by parts:

\begin{equation*} \begin{array}{cc} u=t \amp dv = e^{0.015(T-t)}\,dt\\[1ex] du = dt \amp \displaystyle{v = -\frac{e^{0.015T}}{0.015}e^{-0.015t}} \end{array} \end{equation*}

Thus,

\begin{equation*} \begin{split} \int_0^T te^{0.015(T-t)}\,dt \amp = -\frac{e^{0.015T}}{0.015} t e^{-0.015t}\bigg\vert_0^T + \int_0^T \frac{e^{0.015T}}{0.015}e^{-0.015t} \,dt \\ \amp =-\left[\frac{e^{0.015T}}{0.015} t e^{-0.015t} + \frac{e^{0.015T}}{(0.015)^2}e^{-0.015t}\right]_0^T \\ \amp = \frac{e^{0.015T}}{(0.015)^2} - \frac{T}{0.015} - \frac{1}{0.015^2}. \end{split} \end{equation*}

We also evaluate the integral

\begin{equation*} \int_0^T e^{0.015(T-t)}\,dt = e^{0.015T}\int_0^T e^{-0.015t}\,dt = -\frac{e^{0.015T}}{0.015} e^{-0.015t}\bigg\vert_0^T =-\frac{1}{0.015} + \frac{e^{0.015T}}{0.015} \end{equation*}

Using these results, we have

\begin{equation*} \begin{split} 10,000 \amp = \int_0^5 f(t) e^{0.015(5-t)}\,dt = \int_0^5 (at+b)e^{0.015(5-t)}\,dt \\ \amp = a\left(\frac{e^{0.015(5)}}{(0.015)^2} - \frac{5}{0.015} - \frac{1}{0.015^2}\right) + b \left(-\frac{1}{0.015} + \frac{e^{0.015(5)}}{0.015}\right)\\ \amp \approx 12.8184 a + 5.19228 b. \end{split} \end{equation*}

Similarly, we find

\begin{equation*} 14,000 \approx 25.3805 a + 7.38071 b\text{.} \end{equation*}

Solving this system of equations for \(a\) and \(b\text{,}\) we see that

\begin{equation*} a \approx -29.9991 \text{ and } b \approx 2000\text{.} \end{equation*}

Thus,

\begin{equation*} f(t) \approx -30 t + 2000\text{.} \end{equation*}
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