Precalculus
posted by Ce .
There is a piece of wire 200 cm long. It needs to be cut into 2 pieces, with one piece forming an equilateral triangle and the other piece forming a circle. How long must each piece be to maximize the area of each shape?

area= PI*(L/2)^2 + area of triangle.
Well, it turns out there is a neat formula
area= sqrt(s(sa)(sb)(sc)) where s is the semi perimeter, and a,b,c are sides.
s= 1/2 (3T) where T is triangle side
so areatriangle= sqrt(3T/2)(3T/2T)^3
= sqrt(3T/2 * T^3/8)=sqrt(3T^4/16)
= T^2/4 * sqrt3
so finally, area total=
PI (L/2)^2+ T^2/4 sqrt3
where 3T+PiL= 200cm
or Area=1/4PI* (2003T)^2+T^2/4 sqrt4
check that.
Now with calculus, you take the derivative of area, set to zero, and solve for T, then L.
I assume you have not had this, then plot the parabola, find its maximum, and the corresponding T.
Knowing T, cut three pieces T length, and the remainder Pi*L is the circle perimeter.
check my work.