Problem 53 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

The White House wants to cut the gasoline usage from 140 billion gallons per year in 2007 to 128 billion gallons per year in 2017 . But estimates by the Department of Energy's Energy Information Agency suggest that won't happen. In fact, the agency's projection of gasoline usage from the beginning of 2007 until the beginning of 2017 is given by $$ A(t)=0.014 t^{2}+1.93 t+140 \quad(0 \leq t \leq 10) $$ where \(A(t)\) is measured in billions of gallons/year and \(t\) is in years, with \(t=0\) corresponding to 2007 . a. According to the agency's projection, what will be gasoline consumption at the beginning of \(2017 ?\) b. What will be the average consumption/year over the period from the beginning of 2007 until the beginning of 2017 ?

Step-by-Step Solution

Answer

a. According to the agency's projection, the gasoline usage at the beginning of 2017 will be \(160.7\) billion gallons per year. b. The average gasoline consumption per year over the period from the beginning of 2007 until the beginning of 2017 is \(283.2\) billion gallons per year.

Step 1: Substitute t=10 in A(t) formula

To find the gasoline usage at the beginning of 2017, we will substitute t = 10 into the formula A(t): $$ A(10)=0.014(10)^{2}+1.93(10)+140 $$

Step 2: Evaluate the expression

Now, calculate the value of A(10) by performing the arithmetic operations: $$ A(10)=0.014\cdot100 + 1.93\cdot10 + 140 = 1.4+19.3+140=160.7 $$ So, according to the agency's projection, the gasoline usage at the beginning of 2017 will be 160.7 billion gallons per year. b. Average gasoline usage per year:

Step 1: Find the integral of A(t) over [0,10]

We will integrate the function A(t) over the interval [0,10] to find the total gasoline usage during the 10 year period: $$ \int_0^{10} (0.014 t^{2}+1.93 t+140) dt $$

Step 2: Evaluate the integral

Now, perform the integration and calculate the total gas consumption: $$ \int_0^{10} (0.014 t^{2}+1.93 t+140) dt = 0.00467t^{3}+0.965t^{2}+140t \Big|_0^{10} $$ $$ = (0.00467(10)^{3}+0.965(10)^{2}+140(10)) - (0.00467(0)^{3}+0.965(0)^{2}+140(0)) $$ $$ = 467 + 965 + 1400 = 2832 $$ So, the total gas consumption during this 10-year period is 2832 billion gallons.

Step 3: Calculate the average gasoline usage per year

Finally, we will find the average gasoline usage per year by dividing the total gasoline usage by the number of years: $$ \text{Average Consumption} = \frac{\text{Total Consumption}}{\text{Number of years}} = \frac{2832}{10} = 283.2 $$ So, on average, the consumption per year over the 10-year period is 283.2 billion gallons/year.

Key Concepts

Quadratic Functions

Quadratic functions are essential when modeling relationships where one quantity varies with the square of another. In a quadratic function, the highest power of the variable is squared, which gives the curve its characteristic U or inverted U shape, known as a parabola. A general quadratic function is written as: \[ f(x) = ax^2 + bx + c \]where:

• \(a\), \(b\), and \(c\) are constants,
• \(a eq 0\) ensures the function is quadratic,
• the sign of \(a\) determines the parabola's direction (upward if \(a > 0\), downward if \(a < 0\)).

In the provided problem, \(A(t) = 0.014t^2 + 1.93t + 140\) is a quadratic function used to predict gasoline usage. Here, \(t\) is time in years, and the coefficients reflect the influence of time on usage. To find specific values like gasoline usage in 2017, substitute the appropriate value of \(t\), then solve.

Definite Integrals

Definite integrals are crucial for calculating the area under a curve over a specific interval. When you integrate a function across an interval \([a, b]\), it outputs a single number representing the cumulative value or area;\[\int_a^b f(x) \, dx\]This integration is particularly helpful in physics and engineering to determine volume, density, and total variation in continuous contexts. In the exercise, integrating \(A(t)\) from 0 to 10, \[\int_0^{10} (0.014t^2 + 1.93t + 140) \, dt\], calculates the total gasoline consumption over ten years. To solve, we find the antiderivative of \(A(t)\) and evaluate it from 0 to 10, which yields the total consumption figure of 2832 billion gallons.

Average Value of a Function

The concept of the average value of a function gives insight into the general behavior of a function over an interval. It is calculated by dividing the definite integral of the function over the interval by the interval's length. This is given by:\[\text{Average value} = \frac{1}{b-a} \int_a^b f(x) \, dx\]This is useful for smoothing out variations and assessing overall trends. In the context of the gasoline consumption projection, dividing the total consumption over 10 years (as found from the definite integral) by the number of years, gives the yearly average usage: \[\frac{2832}{10} = 283.2 \, \text{billion gallons/year}\]This average value allows us to estimate typical consumption in a year over the period, providing a clearer picture of usage patterns.