## 20. Maximizing \( y = 12 + 18x - 5x^2 \) --- ### (a) **Estimating the values from a graph of \( y \)** **Step 1: Sketch the graph** - The equation is a downward-opening parabola (since the coefficient of \( x^2 \) is negative). - The vertex of the parabola represents the maximum point. **Step 2: Estimate the vertex** - The vertex (maximum) of \( y = ax^2 + bx + c \) is at \( x = -\frac{b}{2a} \). - Here, \( a = -5 \), \( b = 18 \). So, \[ x = -\frac{18}{2 \times (-5)} = \frac{18}{10} = 1.8 \] **Explanation:** By sketching or using the formula for the vertex, we estimate the maximum occurs at \( x \approx 1.8 \). --- ### (b) **Finding the values using calculus** **Step 1: Find the derivative and set it to zero** \[ y = 12 + 18x - 5x^2 \] \[ \frac{dy}{dx} = 18 - 10x \] Set derivative to zero: \[ 18 - 10x = \implies x = 1.8 \] **Explanation:** The maximum occurs where the derivative is zero, so \( x = 1.8 \). **Step 2: Find the corresponding value of \( y \)** \[ y = 12 + 18(1.8) - 5(1.8)^2 \] \[ = 12 + 32.4 - 16.2 \] \[ = 28.2 \] **Explanation:** Substitute \( x = 1.8 \) back into the original equation to get the maximum value of \( y \). --- ## **Final Answer** - **Value of \( x \) that maximizes \( y \):** \( x = 1.8 \) - **Corresponding value of \( y \):** \( y = 28.2 \)