Find the area bounded by the region given by:
`A = { (x,y) : (x,y) : (x ^(2))/( 25) + (y ^(2))/( 9) le 1 lt (x)/(5) + (y)/(3)}.`

Answer

To find the area bounded by the region defined by the inequalities \( A = \{ (x,y) : \frac{x^2}{25} + \frac{y^2}{9} \leq 1, \frac{x}{5} + \frac{y}{3} > 1 \} \), we can follow these steps:

### Step 1: Identify the curves
The first inequality \( \frac{x^2}{25} + \frac{y^2}{9} \leq 1 \) represents an ellipse centered at the origin, with semi-major axis 5 along the x-axis and semi-minor axis 3 along the y-axis. The second inequality \( \frac{x}{5} + \frac{y}{3} > 1 \) represents a line.

### Step 2: Find the points of intersection
To find the area bounded by these two curves, we first need to determine the points where they intersect. We set the equality of the line to the ellipse:

\[ \frac{x}{5} + \frac{y}{3} = 1 \]

From this, we can express \( y \) in terms of \( x \):

\[ y = 3 - \frac{3}{5}x \]

Now, we substitute this expression for \( y \) into the ellipse equation:

\[ \frac{x^2}{25} + \frac{(3 - \frac{3}{5}x)^2}{9} = 1 \]

Expanding this, we get:

\[ \frac{x^2}{25} + \frac{(3 - \frac{3}{5}x)^2}{9} = 1 \]

Expanding the square:

\[ (3 - \frac{3}{5}x)^2 = 9 - 2 \cdot 3 \cdot \frac{3}{5}x + \left(\frac{3}{5}x\right)^2 = 9 - \frac{18}{5}x + \frac{9}{25}x^2 \]

Substituting this back into the ellipse equation:

\[ \frac{x^2}{25} + \frac{9 - \frac{18}{5}x + \frac{9}{25}x^2}{9} = 1 \]

Multiplying through by 225 (the least common multiple of 25 and 9):

\[ 9x^2 + 25(9 - \frac{18}{5}x + \frac{9}{25}x^2) = 225 \]

This simplifies to:

\[ 9x^2 + 225 - 90x + 9x^2 = 225 \]

Combining like terms:

\[ 18x^2 - 90x = 0 \]

Factoring out \( 18x \):

\[ 18x(x - 5) = 0 \]

Thus, \( x = 0 \) or \( x = 5 \).

### Step 3: Find corresponding y-values
For \( x = 0 \):
\[ y = 3 - \frac{3}{5}(0) = 3 \]

For \( x = 5 \):
\[ y = 3 - \frac{3}{5}(5) = 0 \]

So the points of intersection are \( (0, 3) \) and \( (5, 0) \).

### Step 4: Set up the integral for area
The area we want to find is the area under the ellipse from \( x = 0 \) to \( x = 5 \) minus the area under the line from \( x = 0 \) to \( x = 5 \).

The equation of the ellipse gives us \( y \) as:

\[ y = 3\sqrt{1 - \frac{x^2}{25}} \]

The equation of the line gives us:

\[ y = 3 - \frac{3}{5}x \]

### Step 5: Calculate the area
The area \( A \) can be calculated using the integral:

\[ A = \int_0^5 \left(3\sqrt{1 - \frac{x^2}{25}} - \left(3 - \frac{3}{5}x\right)\right) dx \]

This simplifies to:

\[ A = \int_0^5 \left(3\sqrt{1 - \frac{x^2}{25}} - 3 + \frac{3}{5}x\right) dx \]

### Step 6: Evaluate the integral
We can break this into two parts:

1. The integral of \( 3\sqrt{1 - \frac{x^2}{25}} \)
2. The integral of \( -3 + \frac{3}{5}x \)

Using the formula for the area of a semicircle, we can find the first integral. The second integral can be computed directly.

### Final Calculation
After computing the integrals, we find the area \( A \) to be:

\[ A = \frac{15\pi}{4} - \frac{15}{2} \]

### Conclusion
The area bounded by the region is:

\[ \text{Area} = \frac{15\pi}{4} - \frac{15}{2} \]