Problem 59 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

The price of a certain commodity in dollars/unit at time \(t\) (measured in weeks) is given by $$ p=18-3 e^{-2 t}-6 e^{-\sqrt{/ 3}} $$ What is the average price of the commodity over the 5 -wk period from \(t=0\) to \(t=5\) ?

Step-by-Step Solution

Answer

The average price of the commodity over the 5-week period from \(t=0\) to \(t=5\) is given by: \[ \bar{p}(t) = \frac{1}{5}\Big[90 - \frac{3}{2}(e^{-10} - 1) - 6\sqrt{3}(e^{-\sqrt{/15}} - 1)\Big] \]

Step 1: Set up the integral formula

We will use the average value formula and the given limits of integration to set up the integral. Here, \(a = 0\), \(b = 5\), and the price function \(p(t) = 18 - 3e^{-2t} - 6e^{-\sqrt{/3}}\). \[ \bar{p}(t) = \frac{1}{5 - 0} \int_0^5 \left(18 - 3e^{-2t} - 6e^{-\sqrt{/3}} \right) dt \]

Step 2: Calculate the integral

Now we need to find the indefinite integral of \(p(t)\) with respect to \(t\). We can do this by integrating each term separately: \[ \int \left(18 - 3e^{-2t} - 6e^{-\sqrt{/3}} \right) dt = 18t - \frac{3}{2}e^{-2t} - 6\sqrt{3}e^{-\sqrt{/3}}t + C \] Note that the constant \(C\) is not relevant when calculating a definite integral since it will cancel out when we subtract the values at the limits of integration.

Step 3: Evaluate the integral at the limits of integration

Using the indefinite integral, we can find the definite integral by substituting the limits of integration: \[ \int_0^5 \left(18 - 3e^{-2t} - 6e^{-\sqrt{/3}} \right) dt = \left[18t - \frac{3}{2}e^{-2t} - 6\sqrt{3}e^{-\sqrt{/3}}t \right]_0^5 \] Now we will substitute the values of \(t = 5\) and \(t = 0\): \[ \Big[18(5) - \frac{3}{2}e^{-2(5)} - 6\sqrt{3}e^{-\sqrt{/3}(5)}\Big] - \Big[18(0) - \frac{3}{2}e^{-2(0)} - 6\sqrt{3}e^{-\sqrt{/3}(0)}\Big] \] Simplify this expression: \[ \Big[90 - \frac{3}{2}e^{-10} - 6\sqrt{3}e^{-\sqrt{/15}}\Big] - \Big[0 - \frac{3}{2}e^0 - 6\sqrt{3}e^0\Big] \]

Step 4: Calculate the average value

Now we have the integral value, and we can plug it back into the average value formula to find the average price of the commodity: \[ \bar{p}(t) = \frac{1}{5}\Big[\left(90 - \frac{3}{2}e^{-10} - 6\sqrt{3}e^{-\sqrt{/15}}\right) - \left(0 - \frac{3}{2}e^0 - 6\sqrt{3}e^0\right)\Big] \] Simplify this expression to find the average value: \[ \bar{p}(t) = \frac{1}{5}\Big[90 - \frac{3}{2}(e^{-10} - 1) - 6\sqrt{3}(e^{-\sqrt{/15}} - 1)\Big] \] Now you can use a calculator to find the numerical value of \(\bar{p}(t)\), which represents the average price of the commodity over the 5-week period from \(t=0\) to \(t=5\).

Key Concepts

Definite Integral

The definite integral is a fundamental concept in calculus, representing the accumulation of quantities, such as areas under a curve, total distance traveled, or, in our case, the total cost over time. When we compute the definite integral of a function, we're essentially adding up an infinite number of infinitesimally small values across the interval from the lower limit, denoted by 'a', to the upper limit, denoted by 'b'.

Mathematically, this is represented as \[\int_a^b f(x)dx\], where 'f(x)' is the function we want to integrate. The process of finding the definite integral involves finding the antiderivative (also known as the indefinite integral) of the function and then applying the Fundamental Theorem of Calculus, which tells us to evaluate the antiderivative at the upper limit and subtract the value of the antiderivative at the lower limit.

In our example with the commodity price, calculating the definite integral of the price function over the 5-week period allows us to determine the total accumulated price, which is a crucial step in finding the average price.

Exponential Functions

Exponential functions are a class of mathematical functions characterized by a constant raised to the power of a variable. The general form of an exponential function is \(y = a \cdot b^x\), where 'a' is a constant that serves as the initial value when 'x' is zero, 'b' is the base of the exponential function, and 'x' is the variable.

In our context, the price function includes terms with the mathematical constant 'e'—which is approximately 2.71828—as the base, which is a common base for natural exponential functions. Natural exponential functions, denoted as \(e^x\), are particularly relevant in real-world applications such as compound interest, population growth, and radioactive decay.

Characteristics of Exponential Functions

• Exponential growth or decay depending on whether the base 'b' is greater than or less than one.
• The function never touches the x-axis but gets infinitely close to it as 'x' goes to negative infinity—this is called an asymptote.
• A positive 'a' value means the function is always positive, whereas a negative 'a' means the function is always negative.

In the price function we are analyzing, the negative exponent indicates a decay factor, which is typical in situations like depreciation or, as in our case, a decreasing commodity price over time.

Average Value of a Function

The average value of a function over a particular interval gives us a single number that represents the central or typical value that the function takes on over that interval. It can be thought of as the 'balanced' value, such that if the function had the same value at every point in the interval, the total would be the same as for the actual function. To find the average value of a continuous function on the interval from 'a' to 'b', we use the formula:

\[\text{Average Value} = \frac{1}{b - a} \int_a^b f(x)dx\]This formula implies that we first integrate the function over the specified interval, which gives us the 'total' value of the function, then we divide by the length of the interval to ‘spread out’ this total evenly across the interval.

In the exercise we are discussing, by using this formula with the given price function, we are calculating the average price of a commodity over a 5-week period. This is done by integrating the price function from week 0 to week 5, then multiplying by the reciprocal of the interval's length, which is 1/5 in this case. This average tells us what the price would be if it remained constant at the same total ‘cost’ throughout the 5-week period.