a) The number of units produced cannot be negative, the break-even point is 135.
b) The vertex does not exist, there is no maximum revenue.
c) The maximum profit is $132,596.
d) The price that maximizes the profit is: 1250(514) = $642,500.
a) To find the break-even points, we need to set the revenue equal to the cost and solve for x. The revenue is the selling price per unit times the number of units sold, which is given by 1250x. The cost is the sum of the fixed cost and the variable cost, which is given by 28000 + (x + 222)x = x^2 + 222x + 28000. Setting the revenue equal to the cost, we get:
1250x = x^2 + 222x + 28000
Simplifying and rearranging, we get:
x^2 + 28x - 22400 = 0
Using the quadratic formula, we get:
x = (-28 ± sqrt(28^2 - 4(1)(-22400))) / (2(1)) = (-28 ± 298) / 2
x = -163 or 135
Since the number of units produced cannot be negative, the break-even point is 135.
b) To find the maximum revenue, we need to find the vertex of the parabola given by the revenue function, which is a quadratic function. The vertex occurs at x = -b/2a, where a and b are the coefficients of the quadratic function. In this case, the quadratic function is given by R(x) = 1250x, so a = 0 and b = 1250. Therefore, the vertex occurs at x = -b/2a = -1250/0 = undefined. Since the vertex does not exist, there is no maximum revenue.
c) 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. We already know that R(x) = 1250x and C(x) = x^2 + 222x + 28000. Therefore, the profit function is: P(x) = 1250x - (x^2 + 222x + 28000) = -x^2 + 1028x - 28000
To find the maximum profit, we need to find the vertex of the parabola given by the profit function. The vertex occurs at x = -b/2a, where a and b are the coefficients of the quadratic function. In this case, the quadratic function is given by P(x) = -x^2 + 1028x - 28000, so a = -1 and b = 1028. Therefore, the vertex occurs at x = -b/2a = -1028/(-2) = 514. Plugging x = 514 into the profit function, we get: P(514) = -514^2 + 1028(514) - 28000 = 132,596
Therefore, the maximum profit is $132,596.
d) To find the price that maximizes the profit, we need to find the corresponding value of x that maximizes the profit. We already know that x = 514 maximizes the profit. To find the price, we can plug x = 514 into the selling price function, which is given by 1250x. Therefore, the price that maximizes the profit is: 1250(514) = $642,500.