HELLP!! calculus
posted by Kay on .
A piece of wire 25 m long is cut into two pieces. One piece is bent into a square and the other is bent into a circle.
(a) How much of the wire should go to the square to maximize the total area enclosed by both figures?
(b) how much of the wire should go to the square to minimize the total area enclosed by both figures?

let the length of side of the square be x
Now, the area of a circle is pi r^2, but we have a circumference c = 2pi*r
r = c/2pi and c = (254x)
a = x^2 + pi * [(254x)/2pi]^2
= x^2 + (254x)^2/4pi
This function has a minimum at x = 25/(4 + pi) = 3.5
area is 21.88
but has no global maximum.
At x=0 the area is just the circle = pi * (25/2pi)^2 = 625/4pi = 49.73
At x= 25/4 the area is just the square = 625/16 = 39.06 
Alternate solution
b)
Let the radius of the circle be r
the we need 2πr for the circle
leaving 252πr for the square, then each side of the square is
(252πr)/4
Area = πr^2 + (252πr)^2 /16
= πr^2 + 625/16  (100/16)πr + (4π^2)/16 r^2
this will graph as a parabola opening upwards, so it will have a minimum area.
d(area)/dr = 2πr  100π/16 + π^2 r/2
= 0 for a max of area
2πr + π^2 r/2 = 100π/16
times 16 , then solving for r
32πr + 8π^2 r = 100π
r(32π + 8π^2) = 100π
r = 100/(32+8π) = 25/(8+2π) appr. 1.7503
then the circle needs 2πr = appr. 10.998
leaving 25 – 10.998 or 14.003 m for the square.
Plugging r = 1.7503 into my area equation gives
Minimum Area = 21.879
test by taking a value of r slightly above and slightly below r = 1.7503
if r = 1.7 , area = 21.893
if r = 1.7503 , area = 21.879
if r = 1.8 , area = 21.893
a) The maximum area is obtained when all the wire is used for the circle and NONE is used for the square
i.e. , max area = (12.5^2)π = 490.87 
how is the derivative equal to that? shouldn't the last part (ð^2 r/2)be pi*2r/2?