before it hits the gr 15. Profit function. The profit, \( \$ y \), of a company can \( b_{e} \) approximated by \( y=-\frac{1}{20} x^{2}+75 x-600 \), where \( x \) is the number of goods sold. (i) Explain the meaning of " -600 " in the equation. (ii) How many goods must be sold in order for the company to yield the maximum profit? What is this maximum profit?

Sure, let's analyze the profit function step by step. **Given Profit Function:** \[ y = -\frac{1}{20}x^{2} + 75x - 600 \] where: - \( y \) = Profit in dollars - \( x \) = Number of goods sold --- ### (i) Meaning of "-600" in the Equation The term **-600** in the profit function represents the **fixed costs** or **initial investment** that the company incurs regardless of the number of goods sold. **Explanation:** - **Fixed Costs:** These are costs that do not change with the level of production or sales. They include expenses like rent, salaries of permanent staff, utilities, and other overheads. - In the profit equation, while the terms involving \( x \) (i.e., \( -\frac{1}{20}x^{2} + 75x \)) represent the variable aspects of profit based on sales, the constant term **-600** accounts for the fixed costs that the company has to pay irrespective of how many goods it sells. **Interpretation:** - If the company sells zero goods (\( x = 0 \)), the profit \( y \) would be: \[ y = -\frac{1}{20}(0)^2 + 75(0) - 600 = -600 \] This indicates a loss of \$600 due to fixed costs when no goods are sold. --- ### (ii) Calculating the Number of Goods for Maximum Profit and the Maximum Profit Value The profit function is a quadratic equation in the form: \[ y = ax^{2} + bx + c \] where: - \( a = -\frac{1}{20} \) - \( b = 75 \) - \( c = -600 \) Since \( a < 0 \), the parabola opens downward, meaning the function has a **maximum** point. **Steps to Find the Maximum Profit:** 1. **Find the Vertex of the Parabola:** The vertex of a parabola defined by \( y = ax^{2} + bx + c \) occurs at: \[ x = -\frac{b}{2a} \] 2. **Calculate the Value of \( x \) for Maximum Profit:** \[ x = -\frac{75}{2 \times (-\frac{1}{20})} \] \[ x = -\frac{75}{- \frac{1}{10}} \] \[ x = 75 \times 10 \] \[ x = 750 \] **Interpretation:** - The company must sell **750 goods** to achieve the maximum profit. 3. **Calculate the Maximum Profit (\( y \)):** Substitute \( x = 750 \) back into the profit function: \[ y = -\frac{1}{20}(750)^2 + 75(750) - 600 \] **Compute Each Term:** - \( -\frac{1}{20}(750)^2 = -\frac{1}{20} \times 562,500 = -28,125 \) - \( 75 \times 750 = 56,250 \) - \( -600 \) remains as is. **Add the Terms Together:** \[ y = -28,125 + 56,250 - 600 \] \[ y = 27,525 \] **Interpretation:** - The **maximum profit** the company can achieve is **\$27,525** when **750 goods** are sold. --- **Summary:** - **(i)** The **-600** represents the company's fixed costs. - **(ii)** To attain the maximum profit of **\$27,525**, the company needs to sell **750 goods**.