Question 1174395: Factory A has produced a product for the common users. The total variable cost per unit is
RM 10. The fixed cost is RM 6 200 in a year and the yearly revenue (in RM) is derived by
function R(x) = R(x) = - 2x^2+260x+500, given that x is the number of productions.
(a) Find the production level for breakeven point of factory A.
(b) State the range of number of productions where the factory A should produce to have profit.
(c) State the range of number of productions when the factory A has a loss more than RM 500.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's break down this problem step-by-step.
**(a) Breakeven Point**
1. **Cost Function:**
* Variable Cost (VC) = 10x
* Fixed Cost (FC) = 6200
* Total Cost (C(x)) = VC + FC = 10x + 6200
2. **Revenue Function:**
* R(x) = -2x² + 260x + 500
3. **Breakeven Point:**
* Breakeven occurs when Revenue (R(x)) = Cost (C(x)).
* -2x² + 260x + 500 = 10x + 6200
* -2x² + 250x - 5700 = 0
* Divide by -2: x² - 125x + 2850 = 0
4. **Solve the Quadratic Equation:**
* We can use the quadratic formula: x = [-b ± √(b² - 4ac)] / 2a
* In this case, a = 1, b = -125, c = 2850.
* x = [125 ± √((-125)² - 4 * 1 * 2850)] / 2
* x = [125 ± √(15625 - 11400)] / 2
* x = [125 ± √4225] / 2
* x = [125 ± 65] / 2
* x1 = (125 + 65) / 2 = 190 / 2 = 95
* x2 = (125 - 65) / 2 = 60 / 2 = 30
* **Answer (a):** The breakeven points are at production levels of 30 units and 95 units.
**(b) Range of Production for Profit**
1. **Profit Function:**
* Profit (P(x)) = R(x) - C(x)
* P(x) = (-2x² + 260x + 500) - (10x + 6200)
* P(x) = -2x² + 250x - 5700
2. **Profit Occurs When P(x) > 0:**
* -2x² + 250x - 5700 > 0
* x² - 125x + 2850 < 0 (divide by -2 and reverse the inequality)
3. **Use the Breakeven Points:**
* We already found the breakeven points at x = 30 and x = 95.
* Since the parabola opens downward (-2x²), the profit is positive between the breakeven points.
* **Answer (b):** The factory should produce between 30 and 95 units to have a profit (30 < x < 95).
**(c) Range of Production for Loss > RM 500**
1. **Loss Function:**
* Loss occurs when P(x) < 0.
* We want to find when P(x) < -500.
2. **Set Up Inequality:**
* -2x² + 250x - 5700 < -500
* -2x² + 250x - 5200 < 0
* x² - 125x + 2600 > 0
3. **Solve the Quadratic Equation:**
* x² - 125x + 2600 = 0
* Use the quadratic formula:
* x = [125 ± √((-125)² - 4 * 1 * 2600)] / 2
* x = [125 ± √(15625 - 10400)] / 2
* x = [125 ± √5225] / 2
* x = [125 ± 72.28] / 2
* x1 = (125 + 72.28) / 2 ≈ 98.64
* x2 = (125 - 72.28) / 2 ≈ 26.36
4. **Determine the Range:**
* Since the parabola opens upward, the inequality is satisfied when x < 26.36 or x > 98.64.
* **Answer (c):** The factory has a loss greater than RM 500 when the number of productions is less than 26.36 units or greater than 98.64 units (x < 26.36 or x > 98.64).
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