83.
Minimizing Marginal Cost The marginal cost of a product can be thought of as the cost of producing one additional unit of output. For example, if the marginal cost of producing the 50th product is $6.20, it cost $6.20 to increase production from 49 to 50 units of output. Suppose the marginal cost C (in dollars) to produce x thousand mp3 players is given by the function
C(x) = x2 – 140x + 7400)
Minimizing Marginal Cost
(See Problem 83).
3.3 #84
The marginal cost C (in dollars) of manufacturing x cell phones (in thousands) is given by
C(x) = 5x2 – 200x + 4000.
(a) How many cell phones should be manufactured to minimize the marginal cost?
(b) What is the minimum marginal cost?
I have to answer the second (a,b) question but use the first as a reference.
(a) To minimize the marginal cost, the derivative of the cost function C(x) must be set equal to zero and solved for x.
C'(x) = 10x - 200 = 0
x = 20
Therefore, 20 thousand cell phones should be manufactured to minimize the marginal cost.
(b) To find the minimum marginal cost, substitute x = 20 into the cost function C(x).
C(20) = 5(20)2 - 200(20) + 4000
C(20) = 4000 - 4000 + 4000
C(20) = 4000
Therefore, the minimum marginal cost is $4000.
To find the minimum marginal cost, we need to find the value of x (the number of cell phones) that minimizes the marginal cost function C(x) = 5x^2 - 200x + 4000.
(a) To find the number of cell phones that should be manufactured to minimize the marginal cost, we need to find the x-value of the vertex of the quadratic function. The vertex represents the minimum point of the parabola.
The x-value of the vertex can be found using the formula: x = -b / (2a), where a and b are the coefficients of the quadratic equation in the form ax^2 + bx + c.
In this case, a = 5 and b = -200. Plugging the values into the formula, we have:
x = -(-200) / (2*5)
x = 40
Therefore, 40 thousand cell phones should be manufactured to minimize the marginal cost.
(b) To find the minimum marginal cost, substitute the value of x found in part (a) into the marginal cost function C(x).
C(40) = 5(40)^2 - 200(40) + 4000
C(40) = 5(1600) - 8000 + 4000
C(40) = 8000 - 8000 + 4000
C(40) = 4000
Therefore, the minimum marginal cost is $4000.
To answer the second question (a,b), we can use the same steps as in the first question to determine the number of cell phones to be manufactured that will minimize the marginal cost and to find the minimum marginal cost.
Step 1: Determine the marginal cost function:
The marginal cost function is given as:
C(x) = 5x^2 – 200x + 4000
Step 2: Find the derivative of the marginal cost function:
To find the minimum of the marginal cost function, we need to find its critical points. These points can be found by taking the derivative of the marginal cost function and setting it equal to zero.
C'(x) = 10x - 200
Step 3: Solve for the critical points:
Set the derivative equal to zero and solve for x:
10x - 200 = 0
10x = 200
x = 20
Step 4: Determine if the critical point is a minimum or maximum:
To determine if the critical point is a minimum or maximum, we can use the second derivative test. Take the second derivative of the marginal cost function:
C''(x) = 10
Since the second derivative is positive, the critical point x=20 corresponds to a minimum.
Step 5: Find the minimum marginal cost:
Substitute the value of x=20 into the marginal cost function to find the minimum marginal cost:
C(20) = 5(20)^2 – 200(20) + 4000
C(20) = 5(400) – 4000 + 4000
C(20) = 2000 – 4000 + 4000
C(20) = 2000
The minimum marginal cost is $2000.
Therefore, the answers to question 84 are:
(a) 20 thousand cell phones should be manufactured to minimize the marginal cost.
(b) The minimum marginal cost is $2000.