Problem 142 - Calculus Volume 1

Question

Consider the production of gold during the California gold rush (1848-1888). The production of gold can be modeled by \(G(t)=\frac{(25 t)}{\left(t^{2}+16\right)},\) where \(t\) is the number of years since the rush began \((0 \leq t \leq 40)\) and \(G\) is ounces of gold produced (in millions). A summary of the data is shown in the following figure. Find when the maximum (local and global) gold production occurred, and the amount of gold produced during that maximum.

Step-by-Step Solution

Answer

Maximum gold production was at \( t = 4 \) years, yielding \( 3.125 \) million ounces.

Step 1: Identify the Function and Its Domain

We start with the function \( G(t) = \frac{25t}{t^2 + 16} \), which models the gold production. The domain of this function is given by \( 0 \leq t \leq 40 \), representing years since 1848.

Step 2: Find the Derivative of the Function

To find the critical points where the maximum could occur, first compute the derivative of \( G(t) \). Use the quotient rule: \( \frac{d}{dt}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \) where \( u = 25t \) and \( v = t^2 + 16 \).Compute derivatives: \( u' = 25 \) and \( v' = 2t \).Substitute the values to get: \[ G'(t) = \frac{25(t^2 + 16) - 25t(2t)}{(t^2 + 16)^2} = \frac{400 - 25t^2}{(t^2 + 16)^2} \].

Step 3: Find Critical Points

Set \( G'(t) = 0 \) to find critical points. This simplifies to:\[ 400 - 25t^2 = 0 \]Solving for \( t \),\[ 25t^2 = 400 \]\[ t^2 = 16 \]\[ t = 4 \] and \( t = -4 \) (discard since \( t \geq 0 \)).The critical point is \( t = 4 \). We also check the endpoints \( t = 0 \) and \( t = 40 \).

Step 4: Evaluate Function at Critical Points and Endpoints

Calculate \( G(t) \) at \( t = 0, 4, \) and \( 40 \).- \( G(0) = \frac{25 \times 0}{0^2 + 16} = 0 \)- \( G(4) = \frac{25 \times 4}{4^2 + 16} = \frac{100}{32} = 3.125 \)- \( G(40) = \frac{25 \times 40}{40^2 + 16} = \frac{1000}{1616} \approx 0.619 \) (After simplifying \( 1000/1616 \).)

Step 5: Determine Maximum Production

Compare the function values at \( t = 0, 4, \) and \( 40 \). The greatest value is at \( t = 4 \) with \( G(4) = 3.125 \) million ounces.

Key Concepts

Derivative

In calculus, a derivative represents how a function changes as its input changes. For any function, the derivative is the function's rate of change or slope at any given point. When dealing with gold production modeled by the function \( G(t) = \frac{25t}{t^2 + 16} \), the derivative helps us determine how gold production is changing over time.
To find the critical points where maximums or minimums might occur, we first compute the derivative of the function. Here, the quotient rule is handy for functions involving a ratio like \( G(t) \).

The quotient rule is: \( \frac{d}{dt}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \).

• Identify \( u = 25t \) and \( v = t^2 + 16 \).
• Find their derivatives: \( u' = 25 \) and \( v' = 2t \).

Substituting back, we derive the expression: \[ G'(t) = \frac{25(t^2 + 16) - 25t(2t)}{(t^2 + 16)^2} = \frac{400 - 25t^2}{(t^2 + 16)^2}. \] This expression tells us how fast and in what direction the gold production is changing.

Critical Points

After finding the derivative, our next task is to pinpoint the critical points. A critical point of a function is where its derivative is zero or undefined. These points are potential candidates for local maxima or minima.
In this case, we set the derivative \( G'(t) = \frac{400 - 25t^2}{(t^2 + 16)^2} \) equal to zero to find:\[ 400 - 25t^2 = 0. \]
Solving for \( t \), we have:

• \( 25t^2 = 400 \)
• \( t^2 = 16 \)
• \( t = 4 \) (not \( t = -4 \) since \( t \geq 0 \))

Thus, the critical point is \( t = 4 \). We also need to evaluate the function at the domain's endpoints, \( t = 0 \) and \( t = 40 \), to identify any possible global extrema.

Quotient Rule

The quotient rule is crucial when taking derivatives of functions that are fractions or quotients. It provides a method for differentiating these complex expressions and is expressed by:\[ \frac{d}{dt}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}, \]where \( u \) and \( v \) are functions of \( t \).
Steps to apply the quotient rule include:

• Differentiate the numerator \( u \) to get \( u' \).
• Differentiate the denominator \( v \) to get \( v' \).
• Substitute \( u' \), \( v \), \( u \), and \( v' \) into the formula.

In our function \( G(t) = \frac{25t}{t^2 + 16} \),

• \( u = 25t \) -> \( u' = 25 \)
• \( v = t^2 + 16 \) -> \( v' = 2t \)

Then we have the derivative:\[ G'(t) = \frac{25(t^2 + 16) - 25t(2t)}{(t^2 + 16)^2}. \] Using the quotient rule ensures that we correctly derive the behavior of the function over time.

Global Maximum

Finding the global maximum involves identifying the highest point over the entire domain of a function—in this case, when \( 0 \leq t \leq 40 \). After computing the derivative and finding critical points, we examine these points along with the domain's endpoints to determine the maximum.
For \( G(t) \), we evaluated:

• \( G(0) = 0 \)
• \( G(4) = 3.125 \)
• \( G(40) \approx 0.619 \) (after calculation)

Comparing these values, the largest occurs at \( t = 4 \), where the production reaches \( 3.125 \) million ounces. Thus, at \( t = 4 \), the gold production reached its global maximum, indicating this was the peak of production during the California gold rush.