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In the world of business mathematics, understanding the revenue function is essential for making informed decisions. The revenue function, typically represented as R(p), expresses the total income generated from selling goods or services. For a company that sells physical products, like Nokia with its cell phones, it's crucial to know how changes in price impact revenue.

The relationship between quantity sold and price can often be represented by a linear equation, as seen in the Nokia example with the formula: R(p) = p(-p + 156). This indicates that the number of phones sold, which is reflected as q in the equation, decreases as the price p increases, a common scenario reflecting the law of demand. The goal when managing a revenue function is to find the price point that strikes the perfect balance between price and quantity to maximize revenue. However, revenue maximization does not always imply profit maximization, as costs play a significant role in determining net profit.

Closely linked to the revenue function is the cost function, symbolized as C(q) or C(p), which presents the total expenses associated with the production and sales of goods. For Nokia, the cost function calculation is relatively straightforward - it's the cost of producing one cell phone, $40 in this case, multiplied by the quantity sold: C(q) = 40q.

This equation must also be understood in the context of the price, as the quantity sold is influenced by the selling price, which makes the cost function indirectly dependent on price as well. Accurately determining the cost function is fundamental to ensuring that the price covers production costs and contributes to overall profitability.

Profit is the ultimate measure of a business's success, and the profit function, denoted as P(p), captures this. It's the difference between the revenue and the costs: P(p) = R(p) - C(p). In practical terms for our Nokia example, we substitute q from the cost function into the revenue function to express it entirely in terms of the price p. Therefore, P(p) = p(-p + 156) - 40(-p + 156).

This function is the centerpiece in deciding the optimal selling price for once we find the price that maximizes the profit function, we would have the best possible outcome under the given conditions. This function includes all costs and not just the cost of production, so it gives a complete picture of financial health regarding the sale of a particular item.

The secret to maximizing profit in a mathematical framework lies in calculus, and specifically, in the derivative of the profit function. By taking the derivative of P(p) with respect to price p and finding where this derivative equals zero, we can identify the critical points, which are potential candidates for price points that maximize profit.

This is Step 4 of the provided solution: P'(p) = d/dp [p(-p + 156) - 40(-p + 156)]. By setting P'(p) to zero, we are essentially finding the price at which the rate of change of profit with respect to price is zero - indicating a maximum or minimum. In the context of a business, we are interested in the price that returns the largest profit, which we determine through further analysis of these critical points.