Problem 58 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

It has been estimated that the total production of oil from a certain oil well is given by $$ T(t)=-1000(t+10) e^{-0.1 t}+10,000 $$ thousand barrels \(t\) yr after production has begun. Determine the year when the oil well will be producing at maximum capacity.

Step-by-Step Solution

Answer

The oil well will be producing at maximum capacity when the production begins (\(t = 0\)).

Step 1: Compute the derivative

First, we will find the derivative of the function \(T(t)\) with respect to time \(t\). Using the product rule and exponential rule, we will compute the derivative of \[ T(t) = -1000(t + 10)e^{-0.1t} + 10000. \] The derivative of \(T(t)\) with respect to time is given by \[ T'(t) = \frac{d}{dt}[-1000(t + 10)e^{-0.1t} + 10000]. \] We can use the product rule for \(-1000(t + 10)e^{-0.1t}\), as it involves the product of two functions of \(t\): \(-1000(t + 10)\) and \(e^{-0.1t}\). The product rule is \[ \frac{d}{dt}[u(t)v(t)] = u'(t)v(t) + u(t)v'(t), \] where \(u(t)\) = \(-1000(t + 10)\) and \(v(t)\) = \(e^{-0.1t}\). Now we compute the derivatives of each of them. We have \[ u'(t) = \frac{d}{dt}[-1000(t + 10)] = -1000, \] and \[ v'(t) = \frac{d}{dt}[e^{-0.1t}] = -0.1e^{-0.1t}. \] Applying the product rule, we get \[ T'(t) = -1000[-0.1e^{-0.1t}] + [-1000(t + 10)](-0.1e^{-0.1t}) = 100e^{-0.1t} - 100(t + 10)e^{-0.1t}. \]

Step 2: Determine the critical points

To find the critical points, we will set the derivative equal to zero and solve for \(t\): \[ 100e^{-0.1t} - 100(t + 10)e^{-0.1t} = 0. \] Notice that each term has a common factor of \(100e^{-0.1t}\). Let's factor out this common factor to simplify the equation: \[ 100e^{-0.1t}(1 - (t + 10)) = 0. \] Since the exponential term cannot be zero, we know the product is only zero when the term in parentheses is zero. So, we can set the term in parentheses to zero and solve for \(t\): \[ 1 - (t + 10) = 0. \]

Step 3: Solve for Maximum Production Year

Now, we solve the equation for \(t\): \[ 1 - (t + 10) = 0 \Rightarrow t + 10 = 1 \Rightarrow t = 1 - 10 \Rightarrow t = -9. \] Since the value of \(t\) represents years after the start of production, it doesn't make sense to have a negative value for the maximum production. However, this result means the oil well is producing at its maximum capacity at the beginning of production, which is \(t = 0\). Thus, the oil well will be producing at maximum capacity when the production begins.

Key Concepts

Derivative

In calculus, a derivative represents the rate at which a function is changing at any given point. In our exercise, we're dealing with a function that models the total production of oil over time:
\[T(t) = -1000(t + 10)e^{-0.1t} + 10000.\]
The goal is to find when this production reaches its peak, which requires calculating the derivative, expressed as \(T'(t)\). This derivative will allow us to identify how production changes over time.

• The derivative gives us insights into increasing or decreasing trends of the function over its domain.
• To find the maximum value, we focus on the points where the derivative is zero, as these are potential peaks.

Understanding derivatives is crucial, as they provide vital information about the function's behavior, leading us to critical points essential for identifying maximum or minimum values.

Critical Points

Critical points are important in calculus because they represent places where the function changes direction, potentially indicating maximum or minimum values. For this task, we compute these by setting the derivative to zero:
\[100e^{-0.1t} - 100(t + 10)e^{-0.1t} = 0.\]
After factoring out common terms, we simplify and solve:
\[100e^{-0.1t}(1 - (t + 10)) = 0.\]

• We disregard the exponential factor \(e^{-0.1t}\) since it's never zero.
• Thus, we solve the resulting simple equation: \(1 - (t + 10) = 0\).

Solving this gives \(t = -9\), but logically interpreting the scenario, the maximum occurs at the start, \(t = 0\). This demonstrates how critical point analysis helps pinpoint key changes in functions.

Product Rule

The product rule is a fundamental tool in calculus used to find the derivative of a product of two functions. In the oil production function, \( T(t) = -1000(t + 10)e^{-0.1t} + 10000 \), the term
\[-1000(t + 10)e^{-0.1t}\]
is a product of two functions of \(t\):

• \( u(t) = -1000(t + 10) \)
• \( v(t) = e^{-0.1t} \)

The product rule helps us find the derivative of this product.

The product rule formula states:
\[\frac{d}{dt}[u(t)v(t)] = u'(t)v(t) + u(t)v'(t).\]Applying it, we find:

• \( u'(t) = -1000 \)
• \( v'(t) = -0.1e^{-0.1t} \)

Inserting these into the formula provides:
\[T'(t) = -1000[-0.1e^{-0.1t}] + [-1000(t + 10)](-0.1e^{-0.1t}).\]This rule simplifies the process of calculating derivatives for more complex functions.