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?
To maximize revenue, we need to determine the quantity that maximizes the product of price and quantity demanded. In this case, we can express the revenue (R) as the product of price (p) and quantity demanded (x):
R = p * x
Given that the demand equation is p = √(75 - 2x), we can substitute this equation into the revenue equation:
R = (√(75 - 2x)) * x
To find the quantity that maximizes revenue, we can take the derivative of R with respect to x and set it equal to zero:
dR/dx = (√(75 - 2x)) * (1/2) * (-2) + (√(75 - 2x)) = 0
Simplifying the equation:
(-√(75 - 2x)) + (√(75 - 2x)) = 0
This equation simplifies to:
0 = 0
Since 0 = 0 is a true statement, it means that the derivative dR/dx has no solution. This indicates that revenue is not maximized at a specific quantity.
However, we can determine the maximum revenue by evaluating the revenue equation for the endpoints of the domain, which is when x = 0 and x = 37.5 (as the demand equation p = √(75 - 2x) is defined for x ≤ 37.5). Substituting these values into the revenue equation:
R(0) = (√75) * 0 = 0
R(37.5) = (√(75 - 2 * 37.5)) * 37.5 ≈ 62.5
Therefore, to maximize revenue, approximately 62 dresses should be made and sold each week.
To maximize revenue, we need to find the quantity of dresses that will result in the highest possible product of price and quantity demanded. In this case, the demand equation is given by p = √(75 - 2x), where p is the price and x is the quantity demanded per week.
To find the quantity that maximizes revenue, we can use the concept of marginal revenue. Marginal revenue represents the derivative of the revenue function with respect to quantity. Since revenue is the product of price and quantity, the revenue function is given by R(x) = xp. Taking the derivative of R(x) with respect to x will give us the marginal revenue function, MR(x).
Let's go step by step to find the marginal revenue function and then determine the quantity that maximizes it:
Step 1: Find the revenue function R(x)
R(x) = xp, where p = √(75 - 2x)
R(x) = x * √(75 - 2x)
Step 2: Find the marginal revenue function MR(x)
MR(x) = dR(x)/dx (derivative of R(x) with respect to x)
To find the derivative, we can use the product rule:
MR(x) = √(75 - 2x) + x * (1/2) * (75 - 2x)^(-1/2) * (-2)
Simplifying the expression above, we get:
MR(x) = √(75 - 2x) - x * (75 - 2x)^(-1/2)
Step 3: Determine the quantity that maximizes MR(x)
To find the quantity that maximizes MR(x), we need to set MR(x) equal to zero and solve for x.
0 = √(75 - 2x) - x * (75 - 2x)^(-1/2)
Solving the equation will give us the value of x. However, it might be difficult to solve it analytically. In such cases, we can use numerical methods or graphing software to find the solution.
Once you find the value of x that satisfies MR(x) = 0, substitute this value back into the demand equation to determine the corresponding price p. The quantity and price obtained will give you the optimal number of dresses to be made and sold each week to maximize revenue.