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Revenue maximization occurs when a company charges a price that leads to the highest possible total revenue, without considering costs and profits. In our problem, this is achieved by selling a mix of quantity and price that elicits maximal immediate sales dollars. To explore this concept, let's start with the derived revenue function:
\[ R = p \times (3200 - 8p) = 3200p - 8p^2 \]This quadratic equation represents how revenue depends on the price per tablet, denoted as \( p \). Quadratic functions like this form a parabola when graphed, and the goal is to find the vertex, which yields the maximum point of the parabola. In practical terms, this means differentiating the revenue function to find where its slope is zero:

• Differentiate: \( R' = 3200 - 16p \)
• Solve \( 3200 - 16p = 0 \) for \( p \)

Maximizing revenue in our example means setting \( p = 200 \), the price that leads to the optimal balance between units sold and revenue generated.

Profit maximization is slightly more complex than revenue maximization since it accounts for the costs involved in selling products. In essence, it is finding a price that brings the greatest difference between total revenue and total costs, thereby ensuring the best possible net profit for the retailer. The profit function is:
\[ P = R - C = (3200p - 8p^2) - (35000 + 120x) \]Here, the task is to simplify and determine how to carry the highest profit margin. After simplification, our problem is boiled down to:
\[ P = -8p^2 + 4160p - 419000 \]To find the optimal price for maximum profit, differentiate the profit function and find where the derivative equals zero:

• Derivative: \( P' = -16p + 4160 \)
• Solve \( 16p = 4160 \)

This calculation shows that setting the price at \( p = 260 \) maximizes profit, a slight increase from the revenue-maximizing price due to the cost considerations.

Linear equations are fundamental tools in understanding the relationships between variables like price and quantity sold in economics. In our exercise, the demand function is a classic linear equation:
\[ x = 3200 - 8p \]This equation implies a direct relationship between tablet prices \( p \) and the number of tablets sold \( x \). In simpler terms:

• The higher the price \( p \), the fewer tablets \( x \) are sold.
• Conversely, lowering the price increases the quantity sold.

The negative sign before the \( 8p \) indicates a decrease in sales per unit rise in price, capturing the inverse relationship typical in demand functions. Understanding how to simplify and apply these equations is crucial, as they allow businesses to project outcomes based on varying prices, and tailor strategies for revenue and profit maximization through adjustments in pricing strategies.