Sketch the region enclosed by 2y=5x^(1/2), y=3, 2y+4x=9. Decide whether to integrate with respect to x or y, and then find the area of the region.

your sketch should look like this:

a small "triangular shape"

I would take horizontal slices, that is with respect to y

you first need the intersection of 2y=5x^(1/2) and 2y+4x=9

5√x = 9-4x
25x = 81 - 72x + 16x^2
16x^2 - 97x + 81 = 0
(16x - 81)(x - 1) = 0 , wow it factored
x = 81/16 , or x = 1
as can be seen from the graph, 81/16 is an extraneous root introduced by squaring, so
x = 1, y = 2.5

so you want to integrate with respect to y from 2.5 to 3

Could u integrate it for us

To sketch the region enclosed by the equations 2y = 5x^(1/2), y = 3, and 2y + 4x = 9, we need to determine the intersection points of these curves and then plot the enclosed area.

First, let's find the intersection points between the curves:

1. 2y = 5x^(1/2)
Divide both sides by 2: y = (5/2)x^(1/2)

2. y = 3

3. 2y + 4x = 9
Subtract 4x from both sides: 2y = 9 - 4x
Divide both sides by 2: y = (9 - 4x)/2

Setting each equation equal to each other, we can find the x-values:
(5/2)x^(1/2) = 3
Squaring both sides: (5/2)^2 * (x^(1/2))^2 = 3^2
25/4 * x = 9
Multiply both sides by 4/25: x = (9 * 4) / 25
Simplifying: x = 36/25

Next, we substitute the x-value back into the equation y = (5/2)x^(1/2) to find the y-value:
y = (5/2)(36/25)^(1/2)
Simplifying: y = (5/2)(6/5)
y = 3

So the intersection point of the first two equations is (36/25, 3).

Now, we need to find the intersection points between the second and third equations:

3 = (9 - 4x)/2
Multiply both sides by 2: 6 = 9 - 4x
Subtract 9 from both sides: -3 = -4x
Divide by -4: x = 3/4

Substituting this x-value back into the equation 2y + 4x = 9 to find the y-value:
2y + 4(3/4) = 9
Simplifying: 2y + 3 = 9
Subtract 3 from both sides: 2y = 6
Divide by 2: y = 3

So the intersection point of the second and third equations is (3/4, 3).

Now that we have identified the intersection points, we can sketch the region enclosed by these curves:

- The curve 2y = 5x^(1/2) is a parabola that opens to the right, passing through the point (36/25, 3).
- The straight line y = 3 is a horizontal line.
- The line 2y + 4x = 9 is a straight line that passes through the points (36/25, 3) and (3/4, 3).

The region enclosed by these curves looks like a triangle with a curved top. The base of the triangle is the segment between (36/25, 3) and (3/4, 3), and the curved top is formed by the parabola.

To find the area of this region, we can integrate with respect to x or y. Since the shape is already defined in terms of x, it is more convenient to integrate with respect to x.

The integral for the area is given by:

A = ∫[x=a to b] (f(x) - g(x)) dx,

where f(x) and g(x) are the equations that bound the region.

In this case, the equations that bound the region are y = (5/2)x^(1/2) and y = 3.

So the area A for the region is:

A = ∫[x=a to b] ((5/2)x^(1/2) - 3) dx.

Now we can proceed with evaluating the integral to find the area.

To sketch the region enclosed by the given equations, we can start by finding the intersection points of the curves.

1) To find the intersection point of the first two curves, we equate them:
2y = 5x^(1/2)
Dividing both sides by 2, we have:
y = (5/2)x^(1/2)

2) To find the intersection point of the last two curves, we equate them:
2y + 4x = 9
Solving for y, we get:
2y = 9 - 4x
Dividing both sides by 2, we have:
y = (9 - 4x)/2

Now, let's find the x-values of the intersection points by setting the y-values from both equations equal to each other:
(5/2)x^(1/2) = (9 - 4x)/2

Squaring both sides to eliminate the square root:
(5/2)^2 * (x^(1/2))^2 = ((9 - 4x)/2)^2
25/4 * x = (9 - 4x)^2/4
25x = (9 - 4x)^2

Expanding and simplifying the equation:
25x = 81 - 72x + 16x^2
16x^2 + 97x - 81 = 0

Now we can solve this quadratic equation for x.

Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a, where a = 16, b = 97, and c = -81.

Evaluating the quadratic formula, we get two x-values:
x = (-97 ± √(97^2 - 4 * 16 * -81)) / (2 * 16)

After calculating the values and simplifying, we find that x ≈ -1.45 and x ≈ 2.31.

To decide whether to integrate with respect to x or y, we should sketch the region to see if it lies in the x or y-axis.

After plotting the equations, we find that the region is enclosed between the x-axis and the curves, so it's better to integrate with respect to x.

Now, let's integrate with respect to x to find the area of the region enclosed:
To find the limits of integration, we consider the x-values at the points of intersection: -1.45 and 2.31.

The area (A) of the region is given by:
A = ∫[x1, x2] (upper curve - lower curve) dx

For the upper curve, we have 2y = 5x^(1/2).
Squaring both sides gives: 4y^2 = 25x.
Solving for y, we get: y = √(25/4)x = (5/2)x^(1/2).

For the lower curve, we have y = 3.

Therefore, the integral becomes:
A = ∫[-1.45, 2.31] [(5/2)x^(1/2) - 3] dx

Evaluating this integral will give us the desired area of the region enclosed.