The dimensions that maximize the area of the corral are 187.5π meters and 112.5π meters.
The perimeter of the corral is 600m, so we have the equation
2r + 2w = 600
where r is the radius of the semicircle and w is the width of the rectangle.
The area of the corral is given by:
A = (1/2)πr² + rw
We want to maximize this area, so we can use optimization techniques. Let's solve the first equation for r:
r = 300 - w
Substituting this into the second equation, we get:
A = (1/2)π(300 - w)² + w(300 - w)
Expanding this equation, we get:
A = 45000 - 450πw - 2w²
To find the maximum value of A, we can take the derivative of A with respect to w and set it equal to zero:
A'(w) = -450π - 4w = 0
Solving for w, we get:
w = 112.5π
Substituting this value of w back into the equation for r, we get:
r = 187.5π