Posted by Kay on Wednesday, November 23, 2011 at 5:35pm.
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 = (25-4x)
a = x^2 + pi * [(25-4x)/2pi]^2
= x^2 + (25-4x)^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 25-2πr for the square, then each side of the square is
(25-2πr)/4
Area = πr^2 + (25-2πr)^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?