Hercules Films is deciding on the price of the video release of its film Bride of the Son of Frankenstein. Marketing estimates that at a price of p dollars, it can sell q = 300,000 − 15,000p copies, but each copy costs $4 to make. What price will give the greatest profit?

STUCK ON THIS PROBLEM PLEASE HELP ME OUT!
THANKS IN ADVANCE :)

Thank you, but i entered in the answer you got (10.20) and it said it was wrong.

To find the price that will give the greatest profit, we need to consider the profit for each possible price and then determine the maximum profit.

Let's start by calculating the profit equation. The profit can be determined by subtracting the cost of making each copy from the revenue obtained from selling the copies.

Profit = Revenue - Cost

Revenue is the product of the price and the number of copies sold:
Revenue = p * q

Cost is the product of the cost per copy and the number of copies sold:
Cost = 4 * q

Therefore, the profit equation becomes:
Profit = (p * q) - (4 * q)

We are given that q = 300,000 - 15,000p copies, so we can substitute this into the profit equation:
Profit = (p * (300,000 - 15,000p)) - (4 * (300,000 - 15,000p))

To find the price that will maximize profit, we need to take the derivative of the profit equation with respect to p and set it equal to zero. This will give us the critical points where the profit is either a maximum or minimum.

Differentiating the profit equation with respect to p:
d(Profit)/dp = 300,000 - 30,000p - 4 * (-15,000) + 60,000p
d(Profit)/dp = 300,000 - 30,000p + 60,000 + 60,000p
d(Profit)/dp = -30,000p + 60,000p + 300,000 + 60,000
d(Profit)/dp = 30,000p + 360,000

Setting the derivative equal to zero:
30,000p + 360,000 = 0
30,000p = -360,000
p = -360,000 / 30,000
p = -12

However, a negative price is not practical, so this solution is not meaningful.

Since we don't have a critical point, we can conclude that the profit is either always increasing or always decreasing. In this case, we need to evaluate the profit at the endpoints of the practical price range.

The price must be greater than zero, so let's consider p = 0 and p = 20,000 (the maximum possible price given the quantity sold equation).

Profit at p = 0:
Profit = (0 * q) - (4 * q) = -4q

Profit at p = 20,000:
q = 300,000 - 15,000 * 20,000 = 0
Profit = (20,000 * q) - (4 * q) = 16,000q

Comparing the profits at both endpoints:
Profit at p = 0: -4q
Profit at p = 20,000: 16,000q

Since p = 20,000 will give a positive profit (16,000q) and there is no critical point where the profit is maximum, we can conclude that the price that gives the greatest profit is $20,000.

To find the price that will give the greatest profit, we can start by determining the profit equation.

The profit is the total revenue minus the total cost. We can find the total revenue by multiplying the price p by the number of copies sold q:

Total Revenue = p * q

The total cost is the cost per copy multiplied by the number of copies sold:

Total Cost = $4 * q

The profit equation can be written as:

Profit = Total Revenue - Total Cost
= p * q - $4 * q
= q(p - $4)

Now, the marketing estimates that q = 300,000 − 15,000p. We can substitute this value into the profit equation:

Profit = (300,000 − 15,000p)(p - $4)

To find the price that gives the greatest profit, we need to maximize this profit equation. We can do this by finding the value of p where the derivative of the profit equation with respect to p equals 0.

Taking the derivative of the profit equation:

d(Profit) / dp = 300,000 - 30,000p - 15,000p + 60,000

Simplifying:

d(Profit) / dp = 360,000 - 45,000p

Setting the derivative equal to 0 and solving for p:

360,000 - 45,000p = 0
45,000p = 360,000
p = 360,000 / 45,000
p = 8

So, the price that will give the greatest profit is $8.

revenue = price * demand = pq = p(300000-15000p)

cost for q copies is 4q
profit = revenue-cost

= p(300000-15000p)-4(300000-1500p)
= -15000p^2 + 306,000p - 1,200,000
maximum profit is thus at
p = 306000/30000 = 10.20