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November 22, 2014

November 22, 2014

Posted by **Sckricks** on Monday, November 18, 2013 at 8:08pm.

How much of the wire should go to the square to minimize the total area enclosed by both figures?

- Calculus -
**Reiny**, Monday, November 18, 2013 at 9:20pmlet the side of the square be x

let the side of the triangle be y

4x + 3y = 40

x = (40-3y)/4

height of triangle from the 30-60-90 triangle ratio = √3/2 y

area of triangle = (1/2)(y)(√3/2 y)

= (√3/4)y^2

area of square = [(40-3y)/4]^2

= (1600 - 240y + 9y^2)/16

= 100 - 15y + (9/16)y^2

A = 100 - 15y + (9/16)y^2 + (√3/4)y^2

dA/dy =-15 + 9/8 y + 2√3/4 y

= 0 for a max/min

√3/2 y + 9/8 y = 15

4√3 y + 9y = 135

y = 135/(4√3+9) = appr 8.476

3y = 25.43 m

then 4x = 14.573 m

appr 14.57 m should go for the square

check my arithmetic and algebra

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