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October 23, 2014

October 23, 2014

Posted by **Christine** on Tuesday, April 15, 2014 at 10:10pm.

(a) How much wire should be used for the square in order to maximize the total area?

= 14 m

(b) How much wire should be used for the square in order to minimize the total area?

= ????

- calculus help please -
**Reiny**, Tuesday, April 15, 2014 at 11:41pmlet each side of the square be x m

let each side of the equilateral triangle be 2x

(that way, the height is √3y, from the ratio of the 30-60-90° triangle)

a) for a max area, you are right, all should be used for the square

b) 4x + 6y = 14

2x + 3y = 7

x = (7-3y)/4 OR y = (7-2x)/3

area = x^2 + (1/2)(2y)(√3y)

= x^2 + √3 y^2

= x^2 + √3 ((7-2x)/3)^2

= x^2 + (√3/9)(49 - 28x + 4x^2)

d(area)/dx = 2x + (√3/9)(-28 + 8x) = 0 for a max of area

2x = √3/9(28 - 8x)

18x = 28√3 - 8√3x

x(18 + 8√3) = 28√3

x = 28√3/(18+8√3) = appr 1.522

need 4 x's for the square,**so 6.09 m for the square**, leaving 7.9 m for the triangle for a minimum total area.

check my arithmetic, should have written it out first.

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