Suppose a company has fixed costs of $45,600 and variable cost per unit of

1/3x + 444 dollars, where x is the total number of units produced. Suppose further that the selling price of its product is 2368 − 2/3x dollars per unit.
(a) Find the break-even points.
(b) Find the maximum revenue.
(c) Form the profit function P(x) from the cost and revenue functions.
(d) Find maximum profit.
(e) What price will maximize the profit?

(a) To find the break-even points, we need to find the value of x where the total revenue (selling price per unit multiplied by the number of units) equals the total cost (fixed cost plus variable cost per unit multiplied by the number of units).

The total revenue is given by: R(x) = (2368 - 2/3x)x

The total cost is given by: C(x) = 45600 + (1/3x + 444)x

Setting R(x) equal to C(x) and solving for x:

(2368 - 2/3x)x = 45600 + (1/3x + 444)x

Multiply both sides by 3 to get rid of the fractions:

3(2368 - 2/3x)x = 3(45600 + (1/3x + 444)x)

Simplifying:

7104 - 2x^2 = 136800 + x + 1332x

Rearranging the equation:

-2x^2 - x + 1332x - 136800 - 7104 = 0

Combine like terms:

-2x^2 + 1331x - 144904 = 0

To solve for x, we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac))/(2a)

In this case, a = -2, b = 1331, and c = -144904.

x = (-1331 ± √(1331^2 - 4(-2)(-144904)))/(2(-2))

Simplifying the equation further:

x = (-1331 ± √(1771561 - 1155216))/(2(-2))

x = (-1331 ± √(615345))/(2(-2))

Now, calculate the square root part:

√(615345) ≈ 784.79

Plug in this value:

x = (-1331 ± 784.79)/(2(-2))

To find the break-even points, we need to solve for x. Since negative units don't make sense here, we can disregard the negative solution.

x = (-1331 + 784.79)/(2(-2))
x = (-546.21)/(-4)
x ≈ 136.55

The break-even point is approximately 136.55 units.

(b) To find the maximum revenue, we need to determine the value of x that maximizes the revenue function. In this case, it would be the vertex of the quadratic equation.

The formula for the x-coordinate of the vertex of a quadratic equation in the form ax^2 + bx + c is given by x = -b/2a.

In this case, a = -2 and b = 1331.

x = -1331/(2(-2))
x = -1331/(-4)
x ≈ 332.75

The maximum revenue occurs at approximately 332.75 units.

(c) The profit function P(x) is given by subtracting the total cost from the total revenue:

P(x) = R(x) - C(x)
P(x) = (2368 - 2/3x)x - (45600 + (1/3x + 444)x)
P(x) = (2368x - (2/3)x^2) - (45600 + (1/3)x^2 + 444x)

Combine like terms:

P(x) = 2368x - (2/3)x^2 - 45600 - (1/3)x^2 - 444x

Simplify further:

P(x) = -1448x - (5/3)x^2 - 45600

(d) To find the maximum profit, we need to determine the value of x that maximizes the profit function. Again, we can use the formula for the x-coordinate of the vertex of a quadratic equation.

In this case, a = -5/3 and b = -1448.

x = -(-1448)/(2(-5/3))
x = 1448/(10/3)
x = 1448 * 3/10
x = 4344/10
x = 434.4

The maximum profit occurs at approximately 434.4 units.

(e) To find the price that will maximize the profit, we need to substitute the value of x = 434.4 into the selling price equation:

Selling price = 2368 - 2/3x
Selling price = 2368 - 2/3 * 434.4

Calculating the selling price:

Selling price ≈ 2368 - 2/3 * 434.4
Selling price ≈ 2368 - 289.6
Selling price ≈ 2078.4

The price that will maximize the profit is approximately $2,078.40.

To solve this problem, we will calculate the break-even points, maximum revenue, profit function, maximum profit, and the price that will maximize the profit. Let's go through each step in detail:

(a) Finding the break-even points:
To find the break-even points, we need to determine the number of units at which the company's total revenue equals its total cost.

Total cost (C) = Fixed costs + Variable cost per unit
C = $45,600 + (1/3x + 444)x

Total revenue (R) = Selling price per unit * Number of units sold
R = (2368 - (2/3)x)x

To find the break-even points, we set R equal to C and solve for x:
(2368 - (2/3)x)x = 45,600 + (1/3x + 444)x

Solving this equation will give us the break-even points.

(b) Finding the maximum revenue:
To find the maximum revenue, we need to maximize the revenue function R(x) = (2368 - (2/3)x)x. This can be done by finding the critical points of the function (where the derivative equals zero) and checking for concavity.

(c) Forming the profit function P(x) from the cost and revenue functions:
The profit (P) is calculated by subtracting the total cost (C) from the total revenue (R). So, the profit function is given by:
P(x) = R(x) - C(x)
P(x) = (2368 - (2/3)x)x - (45,600 + (1/3x + 444)x)

(d) Finding the maximum profit:
To find the maximum profit, we need to maximize the profit function P(x). This can be done by finding the critical points of the function (where the derivative equals zero) and checking for concavity.

(e) Finding the price that maximizes the profit:
The price that maximizes the profit is the same as finding the value of x that maximizes the profit function P(x). Once we find the value of x, we can substitute it back into the selling price equation to find the corresponding price.

To solve parts (a), (b), (c), (d), and (e), we will need to perform calculations and solve equations. It would be best to use a calculator or software to simplify and solve the equations involved.

profit is revenue - cost

break-even is when profit = 0
costs: 45600 + (1/3 x + 444)x = 1/3 x^2 + 444x + 45600
revenue : (2368 - 2/3 x)x = 2368x - 2/3 x^2

now use what you know about parabolas.