91Ó°ÊÓ

In the context of optimization in calculus, a cost function is a crucial tool used to calculate the expenses associated with a specific variable. For the exercise provided, the cost function represents the total fuel cost for a truck traveling a certain distance.
The formula for the cost function is derived by incorporating the fuel consumption rate, distance traveled, and cost per liter of fuel. To express the total fuel cost for a truck traveling 200 km, the exercise sets up the equation:

• Total liters of fuel used = \( 200 \times \frac{r(x)}{100} \)


• Total cost = \( 1.15 \times 200 \times \frac{r(x)}{100} \)

This results in the cost function: \( C(x) = 0.575 \left( \frac{4900}{x} + x \right) \). The purpose of this function is to determine how different speeds affect the total travel cost.

Derivatives are fundamental in calculus to analyze how functions change. They provide the rate of change of a function concerning its variables. In optimization problems, derivatives are used to find points where the function achieves extreme values like minima or maxima.
Applying the derivative to the cost function \( C(x) = 0.575 \left( \frac{4900}{x} + x \right) \) involves using rules such as the power rule and the quotient rule. The first derivative, \( C'(x) = 0.575 \left( -\frac{4900}{x^2} + 1 \right) \), reveals how the cost changes as speed \( x \) changes.
This derivative is then used to find the critical points where cost minimization can occur.

Critical points occur where the derivative of a function is zero or undefined. They are essential for finding the minimum or maximum values of a function, crucial in optimization problems.
To find the critical points for the cost function, set the first derivative \( C'(x) \) to zero. Solving the equation \( -\frac{4900}{x^2} + 1 = 0 \) gives the critical point \( x = 70 \) km/h.
This point suggests that 70 km/h is a stationary speed where the cost could either be minimized or maximized. Further tests, like the second derivative test, are used to confirm the nature of these critical points.

The second derivative test helps determine whether a function's critical point is a minimum, maximum, or point of inflection. It relies on the sign of the second derivative \( C''(x) \).
For the cost function, calculating the second derivative provides \( C''(x) = 0.575 \left( \frac{9800}{x^3} \right) \). Substituting the critical point \( x = 70 \) gives a positive value: \( C''(70) > 0 \).
A positive second derivative at this point indicates that the cost function has a local minimum at 70 km/h, confirming that this speed results in the lowest travel fuel cost.

Mathematical modeling involves creating a mathematical representation to analyze real-world situations. It's a crucial part of problem-solving and understanding complex systems in fields like physics and engineering.
In this exercise, mathematical modeling helps optimize truck travel speed to minimize fuel costs. By modeling the fuel consumption and cost relationship using functions, derivatives, and optimization, it becomes possible to make informed decisions about speed for optimal fuel efficiency.
This task highlights the importance of modeling in identifying effective parameters, ensuring resources like fuel are used most efficiently, minimizing costs, and improving overall performance.