A fertilizer producer finds that it can sell its product at a price of p=300-x dollars per unit
when it produces x units of fertilizer. The total production cost (in dollars) for x units is
C(x)= 20,000+24x+0.5x^2. How many units must be manufactured to maximize the profit?
find the max of
f(x) = x*p(x)-c(x)
= x(300-x)-(20000+24x+1/2 x^2)
= -3/2 x^2 + 276x - 20000
f' = -3x + 276
f'=0 at x=92
since f" < 0, that is a max.
thank you
To determine the number of units that must be manufactured to maximize profit, we first need to find the profit function.
The profit is equal to the revenue minus the production cost. The revenue is the product of the price p and the number of units x, i.e., p * x.
The profit function P(x) is given by:
P(x) = p * x - C(x)
Substituting the given values, we have:
P(x) = (300 - x) * x - (20,000 + 24x + 0.5x^2)
Expanding the equation, we get:
P(x) = 300x - x^2 - 20,000 - 24x - 0.5x^2
Combining like terms, we have:
P(x) = -1.5x^2 + 276x - 20,000
To find the number of units that maximizes the profit, we need to find the value of x that corresponds to the vertex of the parabolic equation P(x).
The x-coordinate of the vertex can be found using the formula:
x = -b / (2a)
In this case, a = -1.5 and b = 276. Plugging in the values, we have:
x = -276 / (2 * -1.5)
x = 92
Therefore, to maximize profit, the fertilizer producer must manufacture 92 units.
To find the number of units that must be manufactured to maximize profit, we need to determine the profit function and then find its maximum value.
The profit is equal to the revenue minus the cost. The revenue is calculated by multiplying the price per unit (p) by the number of units sold (x), and the cost is given by the function C(x). Therefore, the profit function (P(x)) can be expressed as:
P(x) = (300 - x)x - (20,000 + 24x + 0.5x^2)
To maximize profit, we need to find the value of x that maximizes P(x). We can do this by taking the derivative of P(x) with respect to x, setting it equal to zero, and solving for x.
Let's start by taking the derivative of P(x):
P'(x) = (300 - x) - [(20,000 + 24x + 0.5x^2)]'
The derivative of a constant is zero, so we only need to differentiate the terms involving x:
P'(x) = 300 - x - (24 + x)
Simplifying further:
P'(x) = 276 - 2x
Now, set P'(x) equal to zero and solve for x:
276 - 2x = 0
2x = 276
x = 138
Therefore, the number of units that must be manufactured to maximize profit is 138 units.