Problem 18 - Essential Calculus

Question

On a given day, the flow rate \(F\) (in cars per hour) on a congested roadway is \(F=\frac{v}{22+0.02 v^{2}}\) where \(v\) is the speed of the traffic in miles per hour. What speed will maximize the flow rate on the road?

Step-by-Step Solution

Answer

The speed that maximizes the flow rate on the road is 550 miles per hour.

Step 1: Differentiate the function

To find maximum or minimum points, first calculate the derivative of the function with respect to \(v\). The derivative of \(F(v)=\frac{v}{22+0.02 v^{2}}\) using the quotient rule is \(F'(v)=\frac{22-0.04v}{(22+0.02v^2)^2}\)

Step 2: Identify the critical points

Critical points occur where the derivative is zero or undefined. So, set \(F'(v)=0\) and solve for \(v\). This gives \(v=550\) as the only critical point. The value of \(F'(v)\) is well-defined for all real numbers, so there are no additional critical points to consider.

Step 3: Analyze the critical points

To determine whether the critical point is a maximum or minimum, use the second derivative test. Computing \(F''(v)\), we get \(-\frac{0.08(22+0.02v^2)-(44+0.04v)(-0.04v)}{(22+0.02v^2)^3}\). Substituting \(v=550\) into the second derivative, we get \(-0.0000073\). Since this is less than zero, the critical point \(v=550\) is a maximum. In this case, the speed to maximize the flow rate on the road is 550 miles per hour.

Key Concepts

Quotient Rule

When working with functions that involve division, we often use the Quotient Rule to differentiate them efficiently. The Quotient Rule helps us find the derivative of a quotient of two functions. Suppose we have a function expressed as a quotient

• Numerator: \( u(v) \)
• Denominator: \( v(v) \)

The rule states that the derivative \( \left( \frac{u}{v} \right)' \) is calculated as: \[\frac{u'(v) \cdot v(v) - u(v) \cdot v'(v)}{(v(v))^2}\]This rule is essential when dealing with complex functions like \( F(v) = \frac{v}{22 + 0.02 v^2} \).
Here, the numerator \( u(v) = v \) and the denominator \( v(v) = 22 + 0.02 v^2 \). Differentiating both, \( u'(v) = 1 \) and \( v'(v) = 0.04v \). Applying the Quotient Rule, we find the derivative \( F'(v) = \frac{22 - 0.04v}{(22 + 0.02v^2)^2} \).
This formula allows us to determine how the flow rate changes with speed, which is crucial in optimization problems.

Critical Points

Critical points play a vital role in understanding the behavior of functions. They occur at values where the derivative of a function equals zero or is undefined. These points indicate potential locations for local maxima, minima, or inflection points.
In our context, we found the derivative of the flow rate function \( F'(v) = \frac{22 - 0.04v}{(22 + 0.02v^2)^2} \).

• Setting \( F'(v) = 0 \) allows us to solve for the critical point: \( v = 550 \).
• Check for undefined points, which do not exist here as the denominator never becomes zero.

Therefore, \( v = 550 \) is the only critical point we have. Finding critical points is essential to determine where a function may exhibit peak performance, like optimizing traffic flow.

Second Derivative Test

The Second Derivative Test is used to confirm the nature of a critical point. After finding a critical point with the first derivative, this test involves computing the second derivative to evaluate the concavity at these points.
The process is simple and goes as follows:

• Compute the second derivative of the function, \( F''(v) \).
• Evaluate \( F''(v) \) at the critical point. For \( v = 550 \), we found \( F''(550) = -0.0000073 \).

Since \( F''(550) < 0 \), the function is concave down at \( v = 550 \), indicating a local maximum.
This test provides a quick, reliable way to assess the nature of critical points, helping us conclude whether we have found a maximum in contexts like optimizing traffic flow, where understanding the best speed is critical to enhancing road efficiency.