I am doing a norman window optimization problem. The perimeter total is 50. the equation is (2r*((50-2r-(2rpi)/2)/2))+((pir^2)/2). All i need is the largest r in the domain of that equation. If i get that i can calculate the max. I have the smallest r=0. I just need the largest possible value of r.
2007-03-08 14:44:49 · 2 answers · asked by Chris 1 in Science & Mathematics ➔ Mathematics
A Norman window has a semicircle on top of a rectangle, right?
The maximum value for r would be the one that has you spending the entire perimeter on the semicircle. The rectangle will become nothing more that the window sill.
So, that brings us to 2r + (pi)r = 50;
r = 50/(2+pi)
2007-03-08 16:26:30 · answer #1 · answered by Doc B 6 · 0⤊ 0⤋
there might want to be a more convenient way, yet curiously that some "wager & examine" might want to be a thanks to bypass (until eventually you know a few calculus). i might want to %. some lengths and widths of the window to confirm what's going to maximize the realm of the rectangle, then compute the realm of the semi-circle in reaction to my outcomes. the realm of the circle is a function, the bigger that is the further gentle will are available in.
2016-10-17 11:16:49 · answer #2 · answered by ? 4 · 0⤊ 0⤋