1. **State the problem:** Find the area enclosed by the curves given by the equations: $$2y=5\sqrt{x}, \quad y=5, \quad 2y+3x=8$$ 2. **Rewrite each equation in terms of $y$ or $x$ for clarity:** - From $2y=5\sqrt{x}$, we get $$y=\frac{5}{2}\sqrt{x}$$ - The second curve is $$y=5$$ - From $2y+3x=8$, solve for $y$: $$2y=8-3x \implies y=\frac{8-3x}{2}$$ 3. **Find the points of intersection to determine the limits of integration:** - Intersection of $y=5$ and $y=\frac{5}{2}\sqrt{x}$: $$5=\frac{5}{2}\sqrt{x} \implies 5=\frac{5}{2}\sqrt{x}$$ Divide both sides by 5: $$\cancel{5}=\frac{\cancel{5}}{2}\sqrt{x} \implies 1=\frac{1}{2}\sqrt{x}$$ Multiply both sides by 2: $$2=\sqrt{x}$$ Square both sides: $$4=x$$ So, intersection point is at $x=4$, $y=5$. - Intersection of $y=5$ and $y=\frac{8-3x}{2}$: $$5=\frac{8-3x}{2}$$ Multiply both sides by 2: $$10=8-3x$$ Subtract 8: $$10-8=-3x \implies 2=-3x$$ Divide both sides by $-3$: $$\frac{2}{-3}=-x \implies x=-\frac{2}{3}$$ Since $x$ cannot be negative for the first curve (due to square root), this intersection is outside the domain of interest. - Intersection of $y=\frac{5}{2}\sqrt{x}$ and $y=\frac{8-3x}{2}$: Set equal: $$\frac{5}{2}\sqrt{x} = \frac{8-3x}{2}$$ Multiply both sides by 2: $$5\sqrt{x} = 8 - 3x$$ Rearranged: $$5\sqrt{x} + 3x = 8$$ Let $t=\sqrt{x}$, so $x=t^2$: $$5t + 3t^2 = 8$$ Rewrite: $$3t^2 + 5t - 8 = 0$$ Solve quadratic: $$t = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 3 \cdot (-8)}}{2 \cdot 3} = \frac{-5 \pm \sqrt{25 + 96}}{6} = \frac{-5 \pm \sqrt{121}}{6}$$ $$t = \frac{-5 \pm 11}{6}$$ Two solutions: - $$t = \frac{-5 + 11}{6} = 1$$ - $$t = \frac{-5 - 11}{6} = -\frac{16}{6} = -\frac{8}{3}$$ (discard negative since $t=\sqrt{x} \geq 0$) So, $t=1 \implies x=1$. 4. **Determine the area enclosed:** The region is bounded between $x=1$ and $x=4$. - The upper curve is $y=5$. - The lower curve is $y=\frac{5}{2}\sqrt{x}$ from $x=1$ to $x=4$. 5. **Set up the integral for the area:** $$\text{Area} = \int_1^4 \left(5 - \frac{5}{2}\sqrt{x}\right) dx$$ 6. **Calculate the integral:** $$\int_1^4 5 dx = 5x \Big|_1^4 = 5(4) - 5(1) = 20 - 5 = 15$$ $$\int_1^4 \frac{5}{2} \sqrt{x} dx = \frac{5}{2} \int_1^4 x^{1/2} dx = \frac{5}{2} \cdot \frac{2}{3} x^{3/2} \Big|_1^4 = \frac{5}{3} (4^{3/2} - 1^{3/2})$$ Calculate $4^{3/2}$: $$4^{3/2} = (\sqrt{4})^3 = 2^3 = 8$$ So: $$\frac{5}{3} (8 - 1) = \frac{5}{3} \times 7 = \frac{35}{3}$$ 7. **Subtract to find the area:** $$15 - \frac{35}{3} = \frac{45}{3} - \frac{35}{3} = \frac{10}{3}$$ **Final answer:** $$\boxed{\frac{10}{3}}$$ This is the area enclosed by the curves.