A lamp has a cost function of C(x) = 2500 + 10x, where x is the number of units produced per day and C(x) is in dollars. The revenue function for these lamps is R (X)= 18x−.001x^2. At least 100 lamps and no more than 2000 lamps may be produced per day.
a. How many lamps should be produced in order to maximize profit (prove this)?
b. What is the cost, in dollars, of producing the number of lamps that maximizes the profit?
c. What is the revenue, in dollars, of producing the number of lamp that maximizes the profit?
P = R - C
P = -2500 + 8 x -.001 x^2
dP/dx = 8 - .002 x
for dP/dx = 0
x = 8/.002 = 4000
so make all you can, 2000
so do C and R with x = 2000
To find the number of lamps that should be produced in order to maximize profit, we need to find the point where the profit is highest. Profit is calculated by subtracting the cost from the revenue, so we need to find the maximum point of the profit function.
The profit function is given by P(x) = R(x) - C(x), where C(x) is the cost function and R(x) is the revenue function.
a. To find the number of lamps that maximize profit, we need to find the maximum point of the profit function P(x). To do this, we can take the derivative of P(x) with respect to x and set it equal to zero. The critical points will then need to be evaluated to determine the maximum point.
P(x) = R(x) - C(x)
P(x) = (18x - 0.001x^2) - (2500 + 10x)
P(x) = -0.001x^2 + 8x - 2500
Now, let's find the derivative of P(x) with respect to x:
P'(x) = -0.002x + 8
Setting P'(x) equal to zero:
-0.002x + 8 = 0
-0.002x = -8
x = -8 / (-0.002)
x = 4000
Now, we need to evaluate this critical point to confirm that it is indeed the maximum point. We can use the second derivative test by finding the second derivative of P(x) and checking its sign at the critical point.
Taking the second derivative of P(x):
P''(x) = -0.002
Since the second derivative is negative, this confirms that the critical point x = 4000 is indeed the maximum point. Therefore, the optimal number of lamps to produce in order to maximize profit is 4000.
b. To find the cost of producing this number of lamps, we can substitute x = 4000 into the cost function C(x):
C(x) = 2500 + 10x
C(4000) = 2500 + 10(4000)
C(4000) = 2500 + 40000
C(4000) = 42500
So, the cost of producing 4000 lamps that maximizes profit is $42,500.
c. To find the revenue of producing this number of lamps, we can substitute x = 4000 into the revenue function R(x):
R(x) = 18x - 0.001x^2
R(4000) = 18(4000) - 0.001(4000^2)
R(4000) = 72000 - 0.001(16000000)
R(4000) = 72000 - 16000
R(4000) = 56000
So, the revenue of producing 4000 lamps that maximizes profit is $56,000.