Question

Suppose that 430 ft of fencing is used to enclose a corral in the shape of a rectangle with a semicircle whose diameter is a side of the rectangle. (In the figure below, the blue outline represents the fencing.) Find the dimensions of the corral with maximum area.

x=?
y=?

i tried couple of ways but got the wrong answer, i know i have to take the derivative and find the minimum value for x but i don't know how to set it up.

Answer 1

2x+2y=430 (x being the diameter and a side; y being another side) That would be your starting equation. I think :)

Answer 2

i did get that and i got farther than that but then got completely lost

Answer 3

Since we can't see your diagram, your problem is not clear to me.
I will assume you have a rectangle with a semicircle mounted on top of the rectangle.

let the base of the rectangle be 2x, (that way I can use x as the radius of the semicircle)
let the height of the rectangle be y

so first equation:
2x + 2y + (1/2)2π(x) = 430
2x + πx + 2y= 430
y = (430-2x-πx)/2

Area = 2xy + (1/2)πx^2
= 2x (430-2x-πx)/2 + (1/2)πx^2

= 430x - 2x^2 - πx^2 + (1/2)πx^2
= 430x - 2x^2 - (1/2)πx^2

d(area)/dx = 430 - 4x - πx = 0

4x + πx = 430
x = 430/(4+π) = 60.21

sub back into y = ... and state your conclusion