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November 29, 2014

November 29, 2014

Posted by **Tara** on Tuesday, July 3, 2012 at 7:35pm.

- Calculus -
**Reiny**, Tuesday, July 3, 2012 at 8:24pmLet the radius of the circle be r and each side of the square be x

so 2πr + 4x = 12

πr + 2x = 6

x = (6-πr)/2

Area = πr^2 + x^2

= πr^2 + ((6-πr)/2)^2

= πr^2 + (3 - πr/2)^2

d(Area)/dr = 2πr + 2(3-πr/2) (-π/2)

= 2πr - π(3-πr/2)

= 0 for a minimus Area

2πr = π(3-πr/2)

2r = 3-πr/2

4r = 6 - πr

4r+πr = 6

r = 6/(4+π) = appr .840

so we need 2πr or 5.2788 ft for the circle,

leaving 6.721 ft for the square.

**radius of circle = .840 ft**

side of each square = 1.68 ft

check:

2πr + 4x

= 2π(.840) + 4(1.68) = 11.998 , not bad

let r = .8 , then x = 1.743

area = π(.8)^2 + 1.743^2 = 5.049

let r = .9, then x = 1.586

area = π(.9)^2 + (1.586)^2 = 5.06.

our answer

r = .84 , x = 1.68

area = π(.84)^ + 1.68^2 = 5.039 , which is lower than either of the

slightly larger and slightly smaller radii.

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