## 20. Maximizing \( y = 12 + 18x - 5x^2 \) --- ### (a) **Estimating the values from a graph of \( y \)** **Step 1: Sketch the graph** - The equation \( y = 12 + 18x - 5x^2 \) is a downward-opening parabola (since the coefficient of \( x^2 \) is negative). - The vertex of a parabola \( y = ax^2 + bx + c \) gives the maximum if \( a < \). - The vertex (maximum) occurs at \( x = -\frac{b}{2a} \). **Explanation**: Estimating from the graph, the vertex is the highest point, so we focus on that. **Step 2: Estimate** - Plug in a few values of \( x \) to see where the maximum occurs: - At \( x = \): \( y = 12 \) - At \( x = 1 \): \( y = 12 + 18 - 5 = 25 \) - At \( x = 2 \): \( y = 12 + 36 - 20 = 28 \) - At \( x = 3 \): \( y = 12 + 54 - 45 = 21 \) - The highest \( y \) value appears around \( x = 2 \). **Final Estimated Answer**: **Maximum at about \( x = 2 \), \( y = 28 \) (estimated from the graph).** --- ### (b) **Finding the values using calculus** **Step 1: Find the critical point** - Take the derivative of \( y \) with respect to \( x \): \[ \frac{dy}{dx} = 18 - 10x \] - Set the derivative to zero to find the maximum: \[ 18 - 10x = \implies x = 1.8 \] **Explanation**: Setting the derivative to zero finds the maximum or minimum. Since the parabola opens downward, this is a maximum. **Step 2: Find the corresponding \( y \) value** - Substitute \( x = 1.8 \) into the original equation: \[ y = 12 + 18(1.8) - 5(1.8)^2 \] \[ = 12 + 32.4 - 5(3.24) \] \[ = 12 + 32.4 - 16.2 \] \[ = 28.2 \] **Final Calculus Answer**: **Maximum at \( x = 1.8 \), \( y = 28.2 \) (exact values from calculus).**

## 20. Maximizing \( y = 12 + 18x - 5x^2 \) --- ### (a) **Estimating the values from a graph of \( y \)** **Step 1: Sketch the graph** - The equation \( y = 12 + 18x - 5x^2 \) is a downward-opening parabola (since the coefficient of \( x^2 \) is negative). - The vertex of a parabola \( y = ax^2 + bx + c \) gives the maximum if \( a < 0 \). - The vertex (maximum) occurs at \( x = -\frac{b}{2a} \). **Explanation**: The vertex is the highest point of the parabola, so we focus on that point to find the maximum value. **Step 2: Estimate** - Plug in a few values of \( x \) to see where the maximum occurs: - At \( x = 0 \): \( y = 12 + 0 - 0 = 12 \) - At \( x = 1 \): \( y = 12 + 18 - 5 = 25 \) - At \( x = 2 \): \( y = 12 + 36 - 20 = 28 \) - At \( x = 3 \): \( y = 12 + 54 - 45 = 21 \) - The highest \( y \) value appears around \( x = 2 \). **Final Estimated Answer**: **Maximum at about \( x = 2 \), \( y = 28 \) (estimated from the graph).** --- ### (b) **Finding the values using calculus** **Step 1: Find the critical point** - Take the derivative of \( y \) with respect to \( x \): \[ \frac{dy}{dx} = 18 - 10x \] - Set the derivative to zero to find the maximum: \[ 18 - 10x = 0 \implies x = 1.8 \] **Explanation**: Setting the derivative to zero identifies critical points, which help determine maximum or minimum values. Given the parabola opens downward, this critical point is a maximum. **Step 2: Find the corresponding \( y \) value** - Substitute \( x = 1.8 \) into the original equation: \[ y = 12 + 18(1.8) - 5(1.8)^2 \] \[ = 12 + 32.4 - 5(3.24) \] \[ = 12 + 32.4 - 16.2 \] \[ = 28.2 \] **Final Calculus Answer**: **Maximum at \( x = 1.8 \), \( y = 28.2 \) (exact values from calculus).** --- ### **Summary of Results** - **Graphical Estimation**: Maximum at \( x \approx 2 \), \( y \approx 28 \) - **Calculus Method**: Maximum at \( x = 1.8 \), \( y = 28.2 \)