suppose the quantity demanded per week of a certain dress is related to the unit price p by the demand equation p=sqrt800-x. write the function that describes the revenue function? To maximize revenue how many dresses should be made and sold each week?
p = √(800-x)
revenue is quantity * price, so
r = x√(800-x)
dr/dt = (1600-3x) / 2√(800-x)
dr/dt=0 when 1600-3x=0, or x=1600/3
a bit of further analysis will show that this is a max, not a min.
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To find the revenue function, we need to multiply the quantity demanded (x) by the unit price (p).
Given that the quantity demanded per week is related to the unit price by the equation p = sqrt(800 - x), we can substitute this expression for p into the revenue function.
So, the revenue function would be:
Revenue = Quantity Demanded * Unit Price
Revenue = x * p
Revenue = x * sqrt(800 - x)
To maximize revenue, we need to find the quantity demanded that maximizes the revenue function.
The easiest way to find the quantity that maximizes the revenue function is by differentiating it with respect to x, setting the derivative equal to zero, and solving for x. However, finding the derivative of the square root function will result in complex calculations.
As an alternative approach, we can use a graph to find the value of x that yields the maximum revenue.
To plot the graph, we need to consider the restrictions on the domain of x. In this case, the quantity demanded cannot be negative, so we have x ≥ 0. Additionally, the quantity demanded cannot exceed the maximum quantity available, which is 800.
By plotting the revenue function for values of x between 0 and 800, we can determine the value of x that yields the maximum revenue.
Unfortunately, we cannot determine the exact value without the specific values for the price and quantity, but by using this approach, you can find the value of x that maximizes revenue for your specific situation.
To find the revenue function, we need to multiply the quantity demanded by the unit price.
Given the demand equation p = sqrt(800 - x), we can express the quantity demanded as:
x = 800 - p^2
Now, to calculate the revenue, we multiply the quantity demanded (x) by the unit price (p):
Revenue = x * p
Substituting the expression for x from the demand equation, we get:
Revenue = (800 - p^2) * p
To maximize revenue, we need to find the value of p that gives the maximum value for the revenue function. We can do this by finding the critical points of the revenue function, which occur where the derivative is equal to zero.
Taking the derivative of the revenue function with respect to p, we get:
dRevenue/dp = 800p - p^3
Setting this derivative equal to zero and solving for p, we get:
800p - p^3 = 0
p(800 - p^2) = 0
This equation has three solutions: p = 0, p = √800, and p = -√800.
Since price cannot be negative in this context, we can discard p = -√800.
Now, to determine which of the remaining two critical points gives the maximum revenue, we can check the second derivative of the revenue function.
Taking the second derivative of the revenue function with respect to p, we get:
d^2Revenue/dp^2 = 800 - 3p^2
Substituting p = √800 into the second derivative, we get:
d^2Revenue/dp^2 = 800 - 3(√800)^2
d^2Revenue/dp^2 = 0
Since the second derivative is zero at p = √800, it does not tell us whether this point is a maximum or minimum. However, since the demand equation p = sqrt(800 - x) implies that price is inversely related to quantity demanded, we can infer that the revenue function is downward sloping. Therefore, the critical point at p = √800 corresponds to a maximum revenue.
To determine the quantity demanded (x) at this critical point, we substitute p = √800 into the demand equation:
x = 800 - (√800)^2
x = 800 - 800
x = 0
Therefore, to maximize revenue, you should make and sell 0 dresses each week. However, this result doesn't seem logical, as it implies no revenue. Hence, it's recommended to review the demand equation and its application to find a more reasonable solution.