To find the dimensions of the rectangle that would maximize its area, we need to differentiate the equation for the area with respect to x and set it equal to zero.
Let's start by differentiating the expression for the area:
A = 2x * sqrt(25 - x^2)
To simplify the differentiation process, let's rewrite the expression as:
A = 2x * (25 - x^2)^(1/2)
Now, using the product rule of differentiation, we can find the derivative of A with respect to x:
dA/dx = 2 * sqrt(25 - x^2) - 2x * (1/2) * (25 - x^2)^(-1/2) * (-2x)
Simplifying this expression further:
dA/dx = 2 * sqrt(25 - x^2) + x^2 / sqrt(25 - x^2)
Setting this derivative equal to zero to find the critical points:
2 * sqrt(25 - x^2) + x^2 / sqrt(25 - x^2) = 0
Multiplying both sides of the equation by sqrt(25 - x^2):
2(25 - x^2) + x^2 = 0
Expanding and simplifying:
50 - 2x^2 + x^2 = 0
2x^2 = 50
x^2 = 25
Taking the square root of both sides and considering only the positive square root:
x = 5
Now that we have found the critical point x = 5, we can substitute this value back into the equation for the area to find the corresponding value of y:
A = 2x * sqrt(25 - x^2)
A = 2 * 5 * sqrt(25 - 5^2)
A = 10 * sqrt(25 - 25)
A = 0
The area of the rectangle at the critical point x = 5 is zero, which does not make sense in this context. Therefore, we need to consider the endpoints of the interval as well.
Since the rectangle is bounded by the x-axis and the semicircle, the possible values for x are in the interval [-5, 5].
First, let's consider the value of x = -5:
A = 2 * (-5) * sqrt(25 - (-5)^2)
A = -10 * sqrt(25 - 25)
A = -10 * sqrt(0)
A = 0
Next, let's consider the value of x = 5:
A = 2 * 5 * sqrt(25 - 5^2)
A = 10 * sqrt(25 - 25)
A = 0
So, at both endpoints of the interval, the area of the rectangle is zero.
Therefore, there is no maximum area for the rectangle within the given domain.