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Understanding the profit function is crucial for businesses looking to maximize their earnings. In the context of mathematics, a profit function represents the relationship between the cost of investment, such as advertising, and the resulting profit. It's typically modeled as a function of the form \( P(x) = a + bx - cx^2 \), where \( P \) denotes the profit, \( x \) is the amount invested, and \( a \), \( b \), and \( c \) are constants representing different aspects of the business scenario.In our exercise, the profit function given is \( P = 230 + 20x - 0.5x^2 \), with profits and expenses in hundreds of dollars. This function indicates that the profit depends linearly on the expenditure up to a certain point, beyond which it decreases due to the quadratic term representing diminishing returns on investment.

Quadratic functions are fundamental in algebra and represent a parabolic graph when plotted. These functions are of the form \( f(x) = ax^2 + bx + c \).The key features of a quadratic function include its vertex, direction of opening (upwards for \( a > 0 \), downwards for \( a < 0 \)), and its axis of symmetry. In the maximum profit problem, we have a quadratic function with a negative leading coefficient (\( -0.5x^2 \)), implying it opens downwards. This is important as it guarantees the existence of a maximum profit point, which is the vertex of the parabola, because a downward-opening parabola has a highest point (its vertex), beyond which the function values start to decrease.

The derivative of a function gives us the slope of the tangent line at any point on the function's graph, which represents the function's rate of change. In economics and business, the derivative application involves using this concept to find maximum or minimum values, which in this case corresponds to maximum profit.To apply derivatives to our profit function, we calculate \( \frac{dP}{dx} = 20 - x \).This derivative tells us how the profit changes with different levels of advertising expenditure. By setting the derivative equal to zero, we can identify the critical points that potentially correspond to maximum or minimum values. Here, the profit changes from increasing to decreasing when the derivative goes from positive to negative, indicating the point of maximum profit.

Critical points are where the derivative of a function is either zero or undefined, often corresponding to the peaks and troughs (maximum and minimum points) on the graph of the function. Locating these points is essential for optimization problems, such as finding the maximum profit.In the exercise, we derived the critical point by setting the first derivative to zero: \( 20 - x = 0 \), which gave us \( x = 20 \). To confirm that this point corresponds to a maximum rather than a minimum or inflection point, we conducted a second derivative test. The second derivative was \( \frac{d^2P}{dx^2} = -1 \), a negative value, which is indicative of a concave down graph at the critical point, thus confirming a maximum. This means that when the company invests $2000 in advertising, it will achieve its maximum profit according to the given profit function.