Question
Suppose that 404 ft of fencing are used to enclose a corral in the shape of a rectangle with a semicircle whose diameter is a side of the rectangle.
Find the dimensions of the corral with maximum area.
x=........ft
Y=......ft
Find the dimensions of the corral with maximum area.
x=........ft
Y=......ft
Answers
rectangle perimter = x + pi*x + 2h
h = (404 - x - pi*x)/2
area = xh + 1/2 pi*(x/2)^2
= x(404 - x - pi*x)/2 + 1/8 pi*x^2
= 202x - x^2/2 - pi*x^2/2 + pi*x^2/8
= -(3pi/8 + 1/2)x^2 + 202x
= x(202 - (3pi/8 + 1/2)x)
This has a maximum when x = 60.187
You can probably get the dimensions from this.
h = (404 - x - pi*x)/2
area = xh + 1/2 pi*(x/2)^2
= x(404 - x - pi*x)/2 + 1/8 pi*x^2
= 202x - x^2/2 - pi*x^2/2 + pi*x^2/8
= -(3pi/8 + 1/2)x^2 + 202x
= x(202 - (3pi/8 + 1/2)x)
This has a maximum when x = 60.187
You can probably get the dimensions from this.