Suppose that each day a company has fixed costs of 400 dollars and variable costs of 0.75x+1430 dollars per unit, where x is the number of units produced that day. Suppose further that the selling price of its product is 1600−0.3x dollars per unit.(a) Each day, the company breaks even at production levels units.
(Enter your answers as a comma-separated list, if necessary)
(b) The maximum daily revenue attained is dollars.
(c) The price that maximizes profit is dollars per unit.
cost c(x) = fixed costs + x*variable costs
revenue r(x) = price * quantity
profit p(x) = revenue - cost
so see what you can do with that.
To find the production level at which the company breaks even, we need to set the total cost equal to the total revenue.
Given:
Fixed costs = 400 dollars
Variable costs = 0.75x + 1430 dollars per unit
Selling price = 1600 − 0.3x dollars per unit
Setting total cost equal to total revenue:
400 + (0.75x + 1430) = (1600 − 0.3x)x
Simplifying the equation:
400 + 0.75x + 1430 = 1600x − 0.3x^2
Combining like terms:
1830 + 0.75x = 1600x − 0.3x^2
Rearranging the equation:
0.3x^2 − 1600x + 1830 + 0.75x = 0
Now, we can use the quadratic formula to solve for x:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 0.3, b = -1600, and c = 1830. Plugging these values into the formula:
x = (-(-1600) ± √((-1600)^2 - 4(0.3)(1830))) / (2(0.3))
Simplifying:
x = (1600 ± √(2560000 - 2196)) / 0.6
Finding the square root:
x = (1600 ± √(2557804)) / 0.6
Calculating the square root:
x = (1600 ± 1600.711) / 0.6
Simplifying:
x = (3200 ± 2672.118) / 0.6
Now we can evaluate both values of x by dividing them by 0.6:
x₁ = (3200 + 2672.118) / 0.6 ≈ 11320.20
x₂ = (3200 - 2672.118) / 0.6 ≈ 878.54
(a) Each day, the company breaks even at production levels of approximately 11320.20 and 878.54 units.
To find the maximum daily revenue attained, we need to find the maximum point on the revenue function. The revenue function can be calculated as:
Revenue = Selling price * Quantity = (1600 - 0.3x)x
To find the maximum point, we can differentiate the revenue function with respect to x and set it equal to zero to find the critical point:
d(Revenue)/dx = -0.3x^2 + 1600x
Setting this equal to zero:
-0.3x^2 + 1600x = 0
Factoring out x:
x(-0.3x + 1600) = 0
Solving for x:
x = 0 or -0.3x + 1600 = 0
From the second equation:
-0.3x = -1600
x = -1600 / -0.3
x = 5333.33
Since negative production levels are not meaningful, we can ignore the solution x = 0. Therefore, the critical point is x = 5333.33.
Substituting this value into the revenue function:
Revenue = (1600 - 0.3(5333.33))(5333.33) ≈ 4,571,111.11
(b) The maximum daily revenue attained is approximately 4,571,111.11 dollars.
To find the price that maximizes profit, we need to find the critical point of the profit function. Profit is calculated as:
Profit = Revenue - Total Cost
Substituting the given equations:
Profit = (1600 - 0.3x)x - (400 + (0.75x + 1430)x)
= (1600x - 0.3x^2) - (400x + 0.75x^2 + 1430x)
= -0.45x^2 + 800x - 400x - 0.75x^2 - 1430x
= -1.2x^2 + 370x
To find the critical point, we can differentiate the profit function with respect to x and set it equal to zero:
d(Profit)/dx = -2.4x + 370
Setting this equal to zero:
-2.4x + 370 = 0
Solving for x:
-2.4x = -370
x = -370 / -2.4
x = 154.17
Since negative production levels are not meaningful, we can ignore the solution x = 0. Therefore, the critical point is x = 154.17.
Substituting this value into the selling price equation:
Price = 1600 - 0.3(154.17)
≈ 1600 - 46.25
≈ 1553.75
(c) The price that maximizes profit is approximately 1553.75 dollars per unit.
To find the answers to these questions, we need to analyze the costs, revenue, and profit of the company given the information provided.
(a) To find the production level where the company breaks even, we need to set the total cost equal to the total revenue. The total cost consists of the fixed costs (400 dollars) and the variable costs (0.75x + 1430 dollars per unit), while the total revenue is the selling price (1600 - 0.3x dollars per unit) multiplied by the number of units produced (x).
Setting the total cost equal to the total revenue, we have the equation:
400 + 0.75x + 1430 = (1600 - 0.3x)x
Simplifying the equation:
0.75x + 1830 = 1600x - 0.3x^2
Rearranging the equation to quadratic form:
0.3x^2 - 0.75x + (1830 - 1600) = 0.3x^2 - 0.75x + 230 = 0
Now we can solve this quadratic equation. Using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 0.3, b = -0.75, and c = 230.
Plugging these values into the quadratic formula:
x = (-(-0.75) ± √((-0.75)^2 - 4 * 0.3 * 230)) / (2 * 0.3)
x = (0.75 ± √(0.5625 - 276)) / 0.6
x = (0.75 ± √(-275.4375)) / 0.6
Since the value inside the square root is negative, there are no real solutions. This means that the company never reaches a break-even point.
(b) To find the maximum daily revenue, we need to maximize the revenue function. The revenue is calculated as the selling price (1600 - 0.3x dollars per unit) multiplied by the number of units produced (x).
So, the revenue function is given by:
R(x) = (1600 - 0.3x)x
To find the maximum point, we can differentiate the revenue function with respect to x and set it equal to zero. Then solve for x.
dR/dx = 1600 - 0.6x = 0
Solving for x:
0.6x = 1600
x = 1600 / 0.6
x = 2666.67
Since the production level cannot be in decimal, we round down to the nearest whole number.
Therefore, the maximum daily revenue is achieved at a production level of 2666 units.
(c) To find the price that maximizes profit, we need to consider the profit function. The profit is calculated as the revenue minus the total cost.
The profit function is given by:
P(x) = R(x) - C(x)
Where R(x) is the revenue function and C(x) is the cost function.
The cost function is given by:
C(x) = 400 + 0.75x + 1430x
Substituting the revenue and cost functions into the profit function:
P(x) = (1600 - 0.3x)x - (400 + 0.75x + 1430x)
Simplifying the equation:
P(x) = 1600x - 0.3x^2 - 400 - 0.75x - 1430x
Combining like terms:
P(x) = -0.3x^2 + (1600 - 0.75 - 1430)x - 400
To find the price that maximizes profit, we need to maximize the profit function. This can be done by differentiating the profit function with respect to x and set it equal to zero.
dP/dx = -0.6x + (1600 - 0.75 - 1430) = 0
Solving for x:
-0.6x + 169.25 = 0
-0.6x = -169.25
x = -169.25 / -0.6
x = 282.08
Since the production level cannot be in decimal, we round down to the nearest whole number.
Therefore, the price that maximizes profit is achieved at a production level of 282 units.