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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 si...Asked by HELP
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.
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.
Answers
Answered by
caroline
2x+2y=430 (x being the diameter and a side; y being another side) That would be your starting equation. I think :)
Answered by
HELP
i did get that and i got farther than that but then got completely lost
Answered by
Reiny
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
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
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