Andrew K. answered • 06/30/19
Student-Athlete and Physics/Computer Science Double Major at MIT
I am not sure what you mean by initial cost, but here is the answer without regarding the initial cost.It seems like the initial cost is likely an added (or subtracted) constant to the profit function, so when we take derivatives, it will go to zero, so the answer will be the same.
The quantity demanded x is inversely proportional to the cube of p, so we can write x = C * p ^ (-3) for some constant C. We can solve for C using the fact that the price is $10 per unit when the quantity demanded is 125, so we can plug in 10 for p and 125 for x. From this we can solve for C, and we get that C = 125,000. Therefore, x = 125,000 * p ^ (-3).
We then use the fact that the total profit equals the profit per unit times the quantity of units demanded. The profit per unit equals the price per unit minus the cost per unit. Therefore, we can write that the total profit = x * (p - 4). Plugging in for x we get that the total profit = 125,000 * (p-4) * p ^ (-3). In order to find the maximum profit, we take the derivative of the profit with respect to the price and set the derivative equal to zero. Solving for p, we get p = 0 or p = 6. Because we know that a price of $0 will not result in maximal profit, we can assume the answer is p = $6. To be sure, we can take the second derivative of the profit with respect to p and plug in p = 6. Doing this, we get that the second derivative is negative at p = 6 which means the profit is concave down and is at a maximum at p = 6. This confirms that at price of $6 gives the maximal profit.
