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A wire of length 12 meters is cut into two parts, one of which is bent into a square and the other into an equilateral triangle. If the sum of their areas is a minimum, find the side length (in meters) of the equilateral triangle. Give your answer correct to two decimal places.

 

 May 1, 2020
 #1
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Let the side length of the equilateral triangle  = S

 

Then the wire  left  for the square perimeter  =  12 - 3S

 

So...the  side of the square  =   [ 12- 3S ] / 4

 

So....the combined areas, A,  of both figures  can be epresented as

 

A = ( [ 12- 3S ] / 4 )^2 +  (√3/4 )S^2

 

A = (1/16) ( 9S^2 - 72S + 144)  + (√3/4)S^2

 

A = (9/16 + √3/4)S^2  - (72/16)S + 9

 

Take the derivative of this and  set to 0

 

A '  = (18/16 + √3/2)S  - (72/16)  =  0

 

A '  =  ( 9/8 + √3/2)S -  4.5  = 0

 

(9/8 + √3/2) S  =  4.5

 

S =  4.5  /  ( 9/8 + √3/2)  ≈  2,26

 

So...the combined area is minimized when the side of the equilateral triangle  ≈ 2.26 m

 

See the graph here to confirm this : https://www.desmos.com/calculator/b81ztsrsgv

 

 

cool cool cool

 May 1, 2020

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