Let's break down the given curves and the steps to find the area enclosed by them.

Step 1: Analyze the curves

1. First curve: $2y = 4\sqrt{x}$

   • Simplify: $y = 2\sqrt{x}$
   • This is the equation of a parabola that opens to the right.

2. Second curve: $y = 5$

   • This is a horizontal line at $y = 5$.

3. Third curve: $2y + 3x = 7$

   • Rearrange for $y$:
     $2y = 7 - 3x \implies y = \frac{7 - 3x}{2}$
   • This is the equation of a line with slope $-\frac{3}{2}$.

Step 2: Sketch the region

To sketch the region, follow these steps:

• Plot the parabola $y = 2\sqrt{x}$. It starts at the origin and opens to the right.
• Plot the line $y = 5$, a horizontal line at $y = 5$.
• Plot the line $y = \frac{7 - 3x}{2}$, a straight line with a negative slope.

Step 3: Find the points of intersection

We need to find where the curves intersect to define the boundaries of the region.

Intersection of $y = 2\sqrt{x}$ and $y = 5$:

$2\sqrt{x} = 5 \implies \sqrt{x} = \frac{5}{2} \implies x = \left(\frac{5}{2}\right)^2 = \frac{25}{4}$ So, the intersection is at $\left(\frac{25}{4}, 5\right)$.

Intersection of $y = \frac{7 - 3x}{2}$ and $y = 5$:

$\frac{7 - 3x}{2} = 5 \implies 7 - 3x = 10 \implies -3x = 3 \implies x = -1$ So, the intersection is at $(-1, 5)$.

Intersection of $y = 2\sqrt{x}$ and $y = \frac{7 - 3x}{2}$:

Set the two equations equal to each other: $2\sqrt{x} = \frac{7 - 3x}{2} \implies 4\sqrt{x} = 7 - 3x$ Square both sides to eliminate the square root: $16x = (7 - 3x)^2 = 49 - 42x + 9x^2$ Simplify: $0 = 9x^2 - 58x + 49$ Solve this quadratic equation using the quadratic formula: $x = \frac{-(-58) \pm \sqrt{(-58)^2 - 4(9)(49)}}{2(9)} = \frac{58 \pm \sqrt{3364 - 1764}}{18} = \frac{58 \pm \sqrt{1600}}{18}$ $x = \frac{58 \pm 40}{18}$ Thus, $x = \frac{98}{18} = \frac{49}{9}$ or $x = \frac{18}{18} = 1$.

So, the two points of intersection are at $x = 1$ and $x = \frac{49}{9}$.

Step 4: Set up the integral

We will integrate with respect to $x$, as the curves are easier to describe in terms of $x$.

The region is bounded by:

• $y = 2\sqrt{x}$ on the bottom,
• $y = \frac{7 - 3x}{2}$ on the top, from $x = 0$ to $x = 1$,
• and $y = 5$ from $x = 1$ to $x = \frac{25}{4}$.

The area is the sum of two integrals:

$\text{Area} = \int_0^1 \left( \frac{7 - 3x}{2} - 2\sqrt{x} \right) dx + \int_1^{\frac{25}{4}} \left( 5 - 2\sqrt{x} \right) dx$

Step 5: Solve the integrals

First integral:

$\int_0^1 \left( \frac{7 - 3x}{2} - 2\sqrt{x} \right) dx$ Simplify and solve.

Second integral:

$\int_1^{\frac{25}{4}} \left( 5 - 2\sqrt{x} \right) dx$ Similarly, simplify and compute.

I'll compute the result for you.The total area of the region enclosed by the curves is approximately $8.17$ square units.

Would you like further details on the steps or any clarifications?

Here are five related questions to explore:

1. How would the area change if the curves were shifted horizontally?
2. What is the impact of rotating one of the curves around an axis?
3. How would you approach finding the volume if this region were revolved around the x-axis?
4. Can the region be bounded differently with the same curves?
5. How would the integral setup change if we integrated with respect to $y$ instead?

Tip: Always ensure you check the boundaries and points of intersection carefully when solving integrals for area.