Suppose the quantity demanded per week of a certain dress is related to price by the demand equation p=√75-2x. In order to maximize revenue, how many dresses should be made and sold each week?

revenue is price * quantity

r = x*p = x√(75-2x)

I hope this is for a calculus class, because we can now use r' to find the max:

r' = 3(25-x)/√(75-2x)
r'=0 when x = 25

To maximize revenue, we need to find the quantity of dresses that will yield the highest value for the product of the quantity and price.

Given the demand equation p = √(75 - 2x), where p represents the price and x represents the quantity demanded per week.

Revenue (R) is calculated by multiplying the price (p) and the quantity demanded per week (x), so R = px.

To maximize revenue, we can either find the critical points of the revenue equation R(x) or maximize the revenue function using calculus. Let's proceed with the calculus approach.

1. Start by expressing revenue as a function of x: R(x) = x * √(75 - 2x).
2. Take the derivative of R(x) with respect to x using the product rule: R'(x) = √(75 - 2x) + x * (-1/2) * (75 - 2x)^(-1/2) * (-2).
3. Simplify the derivative: R'(x) = √(75 - 2x) - x / √(75 - 2x).
4. Set R'(x) equal to zero to find critical points: √(75 - 2x) - x / √(75 - 2x) = 0.
5. Multiply both sides of the equation by √(75 - 2x) to eliminate the denominator: (√(75 - 2x))^2 - x = 0.
6. Simplify the equation: 75 - 2x - x^2 = 0.
7. Rearrange the equation: x^2 + 2x - 75 = 0.
8. Solve the quadratic equation using factoring, completing the square, or the quadratic formula.

In this case, when solving the equation, we find that x = 5 and x = -15. Since the quantity demanded cannot be negative, we discard -15 as a solution.

Therefore, the quantity demanded per week that maximizes revenue is x = 5. So, 5 dresses should be made and sold each week to maximize revenue.