A wire of length L is cut into three pieces, that are bent to form a circle, a s

Question

A wire of length L is cut into three pieces, that are bent to form a circle, a square and an equilateral triangle. How should the cut be made to minimize the sum of the areas of the three shapes?

Explanation / Answer

A wire 300 cm long means ====> perimeter since it has been bent into equilateral triangle and square, their total perimeter is expressed as : perimeter of equilateral triangle = 3a perimeter of square = 4x p = 4x + 3a ====> 300 = 4x + 3a =====> a = [ 300 - 4x ] / [ 3 ] ===> a = [ 100 - (4/3)x ] A of equilateral triangle is (v(3)/4)* a & A of square is x^2 A( total ) = (v(3)/4)* a + x^2 ( lets start substituting : ) A = (v(3)/4) * [ 100 - (4/3)x ] + x^2 A = [ 25v(3) - (1/v(3))x ] + x^2 =====> taking derivative and equal it to zero A ' = - [ 1/v(3) ] + 2x 0 = - [ 1/v(3) ] + 2x [ 1/v(3) ] = 2x ======> x = v(3)/6 = 0.289 cm a = [ 100 - (4/3)*0.289 ] =====> a = 99.6 cm A of equilateral triangle = (v(3)/4)* a = (v(3)/4)* 99.6 = 43.1 cm^2 A of square = x^2 = (0.289)^2 = 0.0835 cm^2 Total Area = 43.1 + 0.0835 = 43.2 cm^2 now logically, when x = 0, we can find max area for equilateral triangle as well as a = 0 for square: x = 0 ====> a = [ 100 - (4/3)*0 ] = 100 cm a = 0 ====> 0 = [ 100 - (4/3)x ] ====> 100 = (4/3)x ====> x = 75 cm we can find two possible maximum areas: A1 (max) = (v(3)/4)* a + x^2 = (v(3)/4)* 100 + 0 = 43.3 cm^2 A2 (max) = (v(3)/4)* a + x^2 = 0 + (75)^2 = 5625 cm^2 i suppose L=300cm

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