First, sketch the region enclosed by the curves \[ 2 y=5 \sqrt{x}, \quad y=3, \quad \text { and } 2 y+3 x=8 \] on your own piece of paper. Decide whether to integrate with respect to \( x \) or \( y \) to compute the area of the region, then find the area. Area:

To find the area of the region enclosed by the curves \( 2y = 5\sqrt{x} \), \( y = 3 \), and \( 2y + 3x = 8 \), we will follow these steps: 1. **Rewrite the equations in a more usable form**: - From \( 2y = 5\sqrt{x} \), we can express \( y \) as: \[ y = \frac{5}{2} \sqrt{x} \] - From \( 2y + 3x = 8 \), we can express \( y \) as: \[ y = 4 - \frac{3}{2}x \] 2. **Find the points of intersection**: - Set \( \frac{5}{2} \sqrt{x} = 3 \) to find the intersection of \( y = \frac{5}{2} \sqrt{x} \) and \( y = 3 \): \[ \frac{5}{2} \sqrt{x} = 3 \implies \sqrt{x} = \frac{6}{5} \implies x = \left(\frac{6}{5}\right)^2 = \frac{36}{25} \] - Set \( \frac{5}{2} \sqrt{x} = 4 - \frac{3}{2}x \) to find the intersection of \( y = \frac{5}{2} \sqrt{x} \) and \( y = 4 - \frac{3}{2}x \): \[ \frac{5}{2} \sqrt{x} + \frac{3}{2}x = 4 \] To solve this, we can square both sides after isolating \( \sqrt{x} \): \[ 5\sqrt{x} + 3x = 8 \implies 5\sqrt{x} = 8 - 3x \] Squaring both sides: \[ 25x = (8 - 3x)^2 \] Expanding and rearranging gives: \[ 25x = 64 - 48x + 9x^2 \implies 9x^2 - 73x + 64 = 0 \] Now we can use the quadratic formula to find \( x \): \[ x = \frac{73 \pm \sqrt{(-73)^2 - 4 \cdot 9 \cdot 64}}{2 \cdot 9} \] 3. **Calculate the area**: - We will integrate with respect to \( y \) since the curves are expressed more conveniently in terms of \( y \). - The area \( A \) can be calculated as: \[ A = \int_{y_1}^{y_2} (x_{\text{right}} - x_{\text{left}}) \, dy \] - The limits of integration will be from \( y = 3 \) to the intersection point found above. Now, let's calculate the intersection points and the area. First, we will solve the quadratic equation \( 9x^2 - 73x + 64 = 0 \) to find the \( x \)-coordinates of the intersections. Let's proceed with the calculations. Solve the quadratic equation by following steps: - step0: Solve by factoring: \(9x^{2}-73x+64=0\) - step1: Factor the expression: \(\left(x-1\right)\left(9x-64\right)=0\) - step2: Separate into possible cases: \(\begin{align}&9x-64=0\\&x-1=0\end{align}\) - step3: Solve the equation: \(\begin{align}&x=\frac{64}{9}\\&x=1\end{align}\) - step4: Rewrite: \(x_{1}=1,x_{2}=\frac{64}{9}\) The solutions to the quadratic equation \( 9x^2 - 73x + 64 = 0 \) yield the following \( x \)-coordinates for the points of intersection: 1. \( x_1 = 1 \) 2. \( x_2 = \frac{64}{9} \approx 7.11 \) Next, we need to find the corresponding \( y \)-values for these \( x \)-coordinates using the equations of the curves. 1. For \( x = 1 \): \[ y = \frac{5}{2} \sqrt{1} = \frac{5}{2} = 2.5 \] (This point is below \( y = 3 \), so we will not use it as a limit for the area.) 2. For \( x = \frac{64}{9} \): \[ y = \frac{5}{2} \sqrt{\frac{64}{9}} = \frac{5}{2} \cdot \frac{8}{3} = \frac{40}{6} = \frac{20}{3} \approx 6.67 \] Now we have the limits of integration: - The lower limit \( y_1 = 3 \) - The upper limit \( y_2 = \frac{20}{3} \) Next, we need to express \( x \) in terms of \( y \) for both curves: - From \( y = \frac{5}{2} \sqrt{x} \): \[ x = \left(\frac{2y}{5}\right)^2 = \frac{4y^2}{25} \] - From \( y = 4 - \frac{3}{2}x \): \[ x = \frac{8 - 2y}{3} \] Now we can set up the integral for the area \( A \): \[ A = \int_{3}^{\frac{20}{3}} \left( \frac{8 - 2y}{3} - \frac{4y^2}{25} \right) dy \] Let's calculate this integral. Evaluate the integral by following steps: - step0: Evaluate using formulas and rules: \(\int_{3}^{\frac{20}{3}} \left(\frac{8-2y}{3}-\frac{4y^{2}}{25}\right) dy\) - step1: Simplify the expression: \(\int_{3}^{\frac{20}{3}} \left(\frac{8}{3}-\frac{2}{3}y-\frac{4y^{2}}{25}\right) dy\) - step2: Evaluate the integral: \(\int \left(\frac{8}{3}-\frac{2}{3}y-\frac{4y^{2}}{25}\right) dy\) - step3: Use properties of integrals: \(\int \frac{8}{3} dy+\int -\frac{2}{3}y dy+\int -\frac{4y^{2}}{25} dy\) - step4: Evaluate the integral: \(\frac{8}{3}y+\int -\frac{2}{3}y dy+\int -\frac{4y^{2}}{25} dy\) - step5: Evaluate the integral: \(\frac{8}{3}y-\frac{y^{2}}{3}+\int -\frac{4y^{2}}{25} dy\) - step6: Evaluate the integral: \(\frac{8}{3}y-\frac{y^{2}}{3}-\frac{4y^{3}}{75}\) - step7: Return the limits: \(\left(\frac{8}{3}y-\frac{y^{2}}{3}-\frac{4y^{3}}{75}\right)\bigg |_{3}^{\frac{20}{3}}\) - step8: Calculate the value: \(-\frac{33209}{2025}\) The integral calculation yields the area \( A \) as: \[ A = -\frac{33209}{2025} \quad \text{or} \quad A = -16\frac{809}{2025} \] Since area cannot be negative, we take the absolute value of the result. Thus, the area of the region enclosed by the curves is: \[ A = \frac{33209}{2025} \quad \text{(approximately 16.4)} \] Therefore, the area of the region enclosed by the curves \( 2y = 5\sqrt{x} \), \( y = 3 \), and \( 2y + 3x = 8 \) is approximately \( 16.4 \) square units.