91Ó°ÊÓ

Quadratic functions are a type of polynomial that are widely used in various mathematical fields. They are generally expressed in the form: \( ax^2 + bx + c \), where \(a\), \(b\), and \(c\) are constants. In this equation, \(x\) represents the variable. The highest exponent of \(x\) is 2, giving it the name 'quadratic'. Quadratic functions graph as parabolas. These can open either upwards or downwards depending on the sign of the coefficient of the \(x^2\) term (\(a\)). If \(a\) is positive, the parabola opens upwards. If \(a\) is negative, it opens downwards.
In the exercise, the revenue function is given by \(R(x) = -x^2 + 50x + 5000\). Here, \(a = -1\), \(b = 50\), and \(c = 5000\). Since our \(a\) is negative, the parabola opens downward. This indicates the revenue function will have a maximum value, not a minimum.

The vertex of a parabola is the point where it reaches its maximum or minimum value. For a quadratic function \( ax^2 + bx + c \), the vertex coordinates \((h, k)\) can be found using the vertex formula. The x-coordinate \( h \) of the vertex is given by:
\[ h = -\frac{b}{2a} \].
In the given problem, we have \(a = -1 \) and \( b = 50 \). Substituting these values into the formula:
\[ h = -\frac{50}{2(-1)} = 25 \].
This tells us that the maximum revenue occurs when there are 25 unsold seats. To find the revenue at this point, we substitute \( x = 25 \) back into the revenue function:
\[ R(25) = -25^2 + 50(25) + 5000 = -625 + 1250 + 5000 = 5625 \].
Thus, the maximum revenue is 5625 dollars.

Optimization is a method in mathematics used to find the best solution from a set of possible choices. In the context of quadratic functions, we often aim to find either the maximum or minimum value by analyzing the function's vertex. This concept is particularly useful in real-world problems where we want to maximize profit, minimize cost, or find the best possible outcome given constraints.
Let's revisit the given problem: A charter bus company wants to maximize revenue. Their revenue function is modeled by a quadratic equation, and since our coefficient of the \(x^2 \) term is negative, we know the function has a maximum point. To find this maximum point, we used the vertex formula to determine the number of unsold seats that maximize revenue and then calculated the corresponding revenue. By substituting \( x = 25 \) into the revenue function, we found that the company gets the highest revenue of 5625 dollars when there are 25 unsold seats. Optimization allowed us to make this discovery straightforwardly.