main algebraic solution

Mathematics
Tutor: None Selected Time limit: 1 Day

May 30th, 2015

Suppose x is bent into a square, so the square will have side length x/4, so its area will be (x/4)^2 = 0.0625x^2 
So 36-x is bent into a rectangle, so 36-x is the perimeter of the rectangle. If the rectangle has length L and width W, then L = 2W, so the perimeter is: 
2(L + W) = 36 - x 
2(2W + W) = 36 - x 
2(3W) = 36 - x 
6W = 36 - x 
W = (36 - x)/6 
The area of the rectangle is: 
R = L*W 
R = 2W * W 
R = 2W^2 
R = 2((36 - x)/6)^2 
R = 2(1296 - 72x + x^2)/36 
R = (1/18)x^2 - 4x + 72 
The total area of the two shapes is: 
A = 0.0625x^2 + (1/18)x^2 - 4x + 72 
A = (17/144)x^2 - 4x + 72 
dA/dx = (17/72)x - 4 
To minimize A, set the derivative to zero: 
(17/72)x - 4 = 0 
(17/72)x = 4 
x = 4 / (17/72) 
x = 4 * (72/17) 
x = 288/17 
W = (36 - (288/17))/6 
W = 324/272 
W = 81/68

May 30th, 2015

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May 30th, 2015
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