A fertilizer producer finds that it can sell its product at a price of p=5000-1.0x dollars per unit when it produces x units of fertilizer. The total production cost (in dollars) for units is c(x)=150000+1250x+0.25x^2
If the production capacity of the firm is at most 1000 units of fertilizer in a specified time, how many units must be manufactured and sold in that time to maximize the profit?
x=? units
1500
800
To maximize profit, we need to find the production level that maximizes the difference between the revenue and the cost.
The revenue is given by the equation:
R(x) = p(x) * x
where p(x) is the price per unit, which is given as p = 5000 - 1.0x.
Substituting in the given value, we have:
R(x) = (5000 - 1.0x) * x = 5000x - x^2
The cost function is given as:
C(x) = 150000 + 1250x + 0.25x^2
The profit function is defined as the difference between the revenue and the cost:
P(x) = R(x) - C(x) = 5000x - x^2 - (150000 + 1250x + 0.25x^2)
We want to find the production level that maximizes the profit:
x = ?
To do that, we need to maximize the profit function P(x). Taking the derivative of P(x) with respect to x and setting it equal to zero gives us the critical points. So, let's find the derivative of P(x):
P'(x) = d/dx (5000x - x^2 - 150000 - 1250x - 0.25x^2)
= 5000 - 2x - 1250 - 0.5x
Setting P'(x) equal to zero and solving for x:
5000 - 2x - 1250 - 0.5x = 0
4250 - 2.5x = 0
-2.5x = -4250
x = -4250 / -2.5
x = 1700
Since the production capacity is at most 1000 units, we need to consider a production level that is within this limit. Therefore, the manufacturer must produce and sell a maximum of 1000 units to maximize the profit.
x = 1000 units
To maximize profit, we need to find the number of units of fertilizer to be manufactured and sold that will yield the maximum difference between revenue and cost.
The revenue earned on selling x units of fertilizer is given by the equation p(x) = 5000 - 1.0x dollars per unit.
The cost of producing x units of fertilizer is given by the equation c(x) = 150000 + 1250x + 0.25x^2 dollars.
The profit, denoted by P(x), is calculated by subtracting the cost function from the revenue function: P(x) = p(x) - c(x).
Let's substitute the given equations into the profit equation to get a simplified expression for the profit:
P(x) = (5000 - 1.0x) - (150000 + 1250x + 0.25x^2)
= 5000 - 1.0x - 150000 - 1250x - 0.25x^2
Now, the production capacity of the firm is at most 1000 units of fertilizer, so the number of units, x, must be less than or equal to 1000.
To find the number of units that maximizes profit, we need to find the critical points of the profit function. The critical points occur where the derivative of the profit function is equal to zero or undefined.
Differentiating the profit function with respect to x:
P'(x) = -1.0 - 1250 - 0.5x
Setting P'(x) equal to zero:
-1.0 - 1250 - 0.5x = 0
Simplifying the equation:
-1251.0 - 0.5x = 0
-0.5x = 1251.0
x = 1251.0 / -0.5
x = -2502
Since x represents the number of units of fertilizer, it cannot be negative. Therefore, we can disregard this solution.
Next, we evaluate the profit function at the endpoints of the production capacity, which are x = 0 and x = 1000.
P(0) = 5000 - 1.0(0) - 150000 - 1250(0) - 0.25(0)^2
= 5000 - 150000
= -145000
P(1000) = 5000 - 1.0(1000) - 150000 - 1250(1000) - 0.25(1000)^2
= 5000 - 1000 - 150000 - 1250000 - 0.25(1000000)
= -1500000
Comparing these values, we see that the profit is maximized at x = 0 (no production or sales), which yields a profit of -145,000 dollars. This result suggests that it may not be economically beneficial to engage in production and sales at the given price and cost functions.
Therefore, based on the given constraints and equations, it is not profitable to manufacture and sell any units of fertilizer in the specified time.