The quantity demanded each month of the Sicard wristwatch is related to the unit price given by the equation below, where p is measured in dollars and x is measured in units of a thousand. To yield a maximum revenue, how many watches must be sold? (Round your answer to the nearest whole number.)
P=41/0.01X^2+1 (0(less than or equal to)x(less than or equal to)20)
? watches
revenue = quantity * price
r(x) = x*p(x)
= 41x/(0.01x^2+1)
dr/dx = (41-.41x^2)/(.01x^2+1)^2
since the denominator is never 0,
max revenue occurs where
(41-.41x^2) = 0
x = 10
Its not wrong, you just need to put in correct units (units of thousands). Its 10,000
that wasn't right
sorry - can you show me my mistake?
idk...it jus said that it was wrong....but when i worked it out i got the same answer as you.
Well, let's see. The quantity demanded is related to the unit price by the equation P = 41/0.01X^2 +1. To maximize revenue, we need to find the number of watches that will yield the highest value for P.
Now, since the equation is a quadratic function, we need to find the vertex of the parabola. The vertex formula tells us that the x-coordinate of the vertex is given by -b/2a. In this case, a = 41/0.01 and b = 0. The x-coordinate of the vertex is therefore -0/2(41/0.01), which is just 0.
So, for maximum revenue, we need to sell 0 watches. But wait a minute... that doesn't make sense, does it? How can you make revenue if you're not selling anything?
Well, in this case, it seems like something's not quite right with the equation or the problem itself. Without further information, it's difficult to determine the exact number of watches that need to be sold to maximize revenue. So, let's just say that you need to sell a few more than 0 watches to make some money!
To find the number of watches that must be sold to yield a maximum revenue, we need to find the value of x that maximizes the revenue function.
The revenue function is given by:
R(x) = p * x
We can substitute the given price equation into the revenue function to get:
R(x) = (41/0.01x^2 + 1) * x
To find the maximum revenue, we need to find the maximum value of the revenue function within the given range of x (0 ≤ x ≤ 20).
To do that, we first need to find the critical points of the revenue function. Critical points occur when the derivative of the function is equal to zero or undefined. So, let's find the derivative of the revenue function:
R'(x) = (d/dx) [(41/0.01x^2 + 1) * x]
R'(x) = (d/dx) [41/0.01x^3 + x]
R'(x) = 41/0.01(3x^2) + 1
Setting the derivative equal to zero and solving for x:
41/0.01(3x^2) + 1 = 0
41/0.01(3x^2) = -1
41/0.03x^2 = -1
x^2 = -1 * (0.03/41)
x^2 = -0.003/41
x^2 = -0.00007317
Since the square of a real number cannot be negative, there are no critical points within the given range. Therefore, there is no maximum revenue within the given range (0 ≤ x ≤ 20).
So, we cannot determine the exact number of watches that must be sold to yield a maximum revenue.