SOLUTION: A Wire Length L Is To Cut Into Two Pieces, One Of Which Is To Be Bent Into The Shape Of A Circle And The Other Into The Shape Of An Equilateral Triangle. Find The Length Of Each Pi

Algebra ->  Equations -> SOLUTION: A Wire Length L Is To Cut Into Two Pieces, One Of Which Is To Be Bent Into The Shape Of A Circle And The Other Into The Shape Of An Equilateral Triangle. Find The Length Of Each Pi      Log On


   

Question 1019815: A Wire Length L Is To Cut Into Two Pieces, One Of Which Is To Be Bent Into The Shape Of A Circle And The Other Into The Shape Of An Equilateral Triangle. Find The Length Of Each Piece So That The Sum Of The Enclosed Areas Is A Minimum
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
R= radius of the circle
x= side of the triangle
L length of the wire before cutting it

2pi%2AR= length of wire used for the circle
3x=L-2pi%2AR= length of wire used for the triangle
3x=L-2pi%2AR<-->x=%28L-2pi%2AR%29%2F3

Area of the triangle =
Area of the circle =pi%2AR%5E2
Total area =y=pi%2AR%5E2%2Bsqrt%283%29%28L-2pi%2AR%29%5E2%2F36
y=pi%2AR%5E2%2Bsqrt%283%29%28L%5E2-4L%2Api%2AR%2B4pi%5E2%2AR%5E2%29%2F36



That is a quadratic function.
Quadratic functions of the form y=ax%5E2%2Bbx%2Bc with a%3E0
have a minimum at x=%28-b%29%2F2a .
In the case of ,
x=R , a=%28pi%2Bsqrt%283%29pi%5E2%2F9%29%3E0 , and b=-sqrt%283%29L%2Api%2F9 ,
so the minimum is at