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A demand function is a mathematical representation that describes the relationship between the price of a product and the quantity of the product that consumers are willing to buy. In this problem, the demand function provided was for corn, and it was given as:\[ p = \frac{6,570,000}{q^{1.3}} \]Here, \(p\) represents the price per bushel of corn, and \(q\) is the number of bushels sold. This equation indicates that the price per bushel decreases as the quantity sold increases. It reflects the typical economic principle of demand: as more units are sold, the price tends to fall.

• The higher the exponent in \(q\), the steeper the curve representing the demand.
• The constants like 6,570,000 determine the scale of the demand relation.

Thus, understanding this function allows farmers to predict how changes in price could impact the number of bushels they sell.

Calculating a derivative is a critical step in understanding how a function behaves, particularly to determine its rate of change. In calculus, the derivative functions as a tool to show how one quantity changes in relation to another. For revenue optimization, finding the derivative of the revenue function helps reveal when revenue stops increasing and starts decreasing. The formula to find the first derivative of a function, generally denoted as \( f'(x) \), uses the power rule:\[ R'(q) = -\frac{6,570,000 \times 0.3}{q^{1.3}} \]Here's how the process looks:

• Take the exponent 0.3, multiply it with the constant, resulting in the first part \(-6,570,000 \times 0.3\).
• Subtract one from the original exponent of \(q^{0.3}\), this becomes \(q^{1.3}\).

Analyzing this formula helps determine whether the revenue is increasing, decreasing, or at a standstill.

Critical points in calculus are values of the variable where the derivative is zero or undefined, indicating potential maximum or minimum points on the graph of a function. However, in this exercise, the derivative \( R'(q) \) was never zero, meaning the critical point had to be determined differently:

• Instead, the critical analysis focused on the interval \([10,000, \infty)\).
• Since \( R'(q) \) is negative for \( q = 10,000 \), the revenue function is decreasing at the beginning of the interval.

This shows that the optimum situation happens at the minimal sales figure applicable, \( q = 10,000 \), as a result of decreasing revenue beyond this.

To find the maximum revenue, combining the mechanisms of both the demand function and its derivative is essential. This process allows for pinpointing exactly at what quantity the revenue peaks before declining:

• By finding \( q = 10,000 \) from the exploration of critical points, the maximum revenue was calculated.
• Using the demand function at this sales level determines the price \( p = 2.09 \) per bushel.
• Multiplying \( q \) by \( p \) gives the highest revenue at this point: \( 20,934,460 \).

Identifying this maximum figure not only ensures the farmers can strategize their sales approach but also guarantees they are operating at optimal efficiency in terms of pricing and quantity of corn sold.