Suppose a company has fixed costs of $38,400 and variable cost per unit of
1
3
x + 222 dollars,
where x is the total number of units produced. Suppose further that the selling price of its product is
1454 −
2
3
x dollars per unit.
(a) Find the break-even points. (Enter your answers as a comma-separated list.)
x =
(b) Find the maximum revenue. (Round your answer to the nearest cent.)
$
(c) Form the profit function P(x) from the cost and revenue functions.
P(x) =
Find maximum profit.
$
(d) What price will maximize the profit? (Round your answer to the nearest cent.)
$
its 1/3x+222 and 2/3x
To calculate the break-even points, we need to find the values of x for which the revenue is equal to the total costs.
The revenue function is given by:
R(x) = (1454 - (2/3)x)x
The cost function can be expressed as:
C(x) = 38400 + (1/3)x + 222x
To find the break-even points, set R(x) equal to C(x) and solve for x.
1454 - (2/3)x = 38400 + (1/3)x + 222x
Let's solve this equation step by step:
1. Combine like terms on the right side:
1454 - (2/3)x = 38400 + (223/3)x
2. Bring all the x terms to one side:
1454 - (2/3)x - (223/3)x = 38400
3. Combine the x terms on the left side:
1454 - (2/3 - 223/3)x = 38400
4. Simplify the coefficients of x:
1454 - (225/3)x = 38400
5. Multiply both sides of the equation by 3 to get rid of the fraction:
4362 - 225x = 115200
6. Subtract 4362 from both sides of the equation:
-225x = 110838
7. Divide both sides by -225:
x = -110838/225
The break-even point is approximately -492.96 units.
Since producing a negative number of units is not possible, we can conclude that there is no break-even point.
To find the maximum revenue, we can use the revenue function R(x) = (1454 - (2/3)x)x.
To find the maximum, we take the derivative of R(x) with respect to x and set it equal to zero.
R'(x) = 1454 - (2/3)x - (2/3)x = 1454 - (4/3)x
To find the critical point, set R'(x) equal to zero and solve for x:
1454 - (4/3)x = 0
Multiply both sides by 3 to eliminate the fraction:
4362 - 4x = 0
Subtract 4362 from both sides:
-4x = -4362
Divide both sides by -4:
x = 4362/4
x = 1090.5
Since the number of units can only be a whole number, the maximum revenue occurs at x = 1090 units.
To calculate the profit function, we subtract C(x) from R(x):
P(x) = R(x) - C(x) = (1454 - (2/3)x)x - (38400 + (1/3)x + 222x)
Now let's simplify:
P(x) = (1454x - (2/3)x^2) - (38400 + (1/3)x + 222x)
P(x) = 1454x - (2/3)x^2 - 38400 - (1/3)x - 222x
Combine like terms:
P(x) = - (2/3)x^2 + (1454 - (1/3) - 222)x - 38400
Simplify further:
P(x) = - (2/3)x^2 + (1231 - 223)x - 38400
To find the maximum profit, we need to find the value of x that maximizes the profit function P(x). We can do this by taking the derivative of P(x) with respect to x and setting it equal to zero.
P'(x) = - (4/3)x + (1231 - 223) = 0
Simplify:
-(4/3)x + 1008 = 0
Multiply both sides by -3 to eliminate the fraction:
4x - 3024 = 0
Add 3024 to both sides:
4x = 3024
Divide both sides by 4:
x = 756
The maximum profit occurs at x = 756 units.
To find the price that maximizes profit, we substitute the value of x = 756 into the selling price function:
S(x) = 1454 - (2/3)x
S(756) = 1454 - (2/3)(756)
Simplify:
S(756) = 1454 - 504
S(756) = 950
The price that maximizes profit is $950.
To solve this problem, we need to use the concepts of cost, revenue, and profit.
Let's start by finding the break-even points, which occur when the revenue equals the total cost.
The cost function is the sum of fixed costs ($38,400) and the variable cost per unit, which is given by 13x + 222.
The revenue function is the selling price per unit (1454 - (2/3)x) multiplied by the total number of units produced (x).
(a) To find the break-even points, we set the revenue equal to the cost and solve for x:
1454 - (2/3)x = 38,400 + 13x + 222
Combine like terms:
1454 - 38,400 - 222 = 13x + (2/3)x
-36,168 = (25/3)x
To isolate x, multiply both sides by 3/25:
x = (-36,168) * (3/25)
So, the break-even point is x ≈ -4,340.672.
However, it doesn't make sense to have a negative number of units produced, so we ignore this answer.
Therefore, there is no break-even point in this case (the company is always operating at a loss).
(b) To find the maximum revenue, we need to determine the maximum value of the revenue function.
The revenue function is given by:
Revenue (R) = (1454 - (2/3)x) * x
To find the maximum revenue, we can use techniques from calculus. Take the derivative of the revenue function with respect to x, set it equal to zero, and solve for x.
R'(x) = (1454 - (2/3)x) + x(-2/3) = 0
Simplifying:
1454 - (2/3)x - (2/3)x = 0
Combine like terms:
1454 - (4/3)x = 0
To isolate x, divide both sides by (4/3):
x = (3/4) * 1454
x ≈ 1089.75
So, the maximum revenue occurs at x ≈ 1089.75 units.
(c) The profit function (P) is obtained by subtracting the cost function from the revenue function:
P(x) = R(x) - C(x)
P(x) = (1454 - (2/3)x) * x - (38,400 + 13x + 222)
Simplifying:
P(x) = 1454x - (2/3)x^2 - 38,400 - 13x - 222
P(x) = (-2/3)x^2 + 1441x - 38,622
(d) To find the price that maximizes profit, we need to find the value of x that leads to the maximum profit.
The profit function is given by P(x) = (-2/3)x^2 + 1441x - 38,622.
Again, we can use calculus to find the maximum. We take the derivative of the profit function with respect to x, set it equal to zero, and solve for x.
P'(x) = (-4/3)x + 1441 = 0
Simplifying:
(-4/3)x = -1441
To isolate x, divide both sides by (-4/3):
x = (-1441) * (-3/4)
x ≈ 1,080.75
Therefore, to maximize profit, the company should produce approximately 1,080.75 units.
To find the price that maximizes profit, plug this value of x into the revenue function:
Price = 1454 - (2/3)(1,080.75)
Price ≈ $727.17
So, the price that maximizes profit is approximately $727.17.