SOLUTION: Take any wire of length 16 inches. Cut the wire into two pieces such that the sum of the areas of the square and circle formed by the pieces of wire is the minimum.
A = Area o
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A = Area o
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Question 976989: Take any wire of length 16 inches. Cut the wire into two pieces such that the sum of the areas of the square and circle formed by the pieces of wire is the minimum.
A = Area of square + Area of circle.
= (S/4)^2 + C^2/4(pi symbol) where C = 2(pi symbol)r
1) S + C =____________
2) A =_______ + _______________ (Write the equation in terms of C. )
3) Find dA/dC.
4) Set dA/dC = 0 and solve for C.
5) Check if A’’ is positive. If it is positive, the total area is minimum.
6) S = ___________ and C =_______________
Please help me with this problem. Thank you so much!! Found 2 solutions by Alan3354, htmentor:Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! Take any wire of length 16 inches. Cut the wire into two pieces such that the sum of the areas of the square and circle formed by the pieces of wire is the minimum.
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s = side length of the square.
Circumference of the circle = 16 - 4s = 2pi*r
r = (16-4s)/2pi = (8-2s)/pi
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Area of circle = pi*(8 - 2s)^2
Area of square = s^2
Total = s^2 + pi*(8 - 2s)^2
dA/ds = 2s + pi*2*(8-2s)*(-2) = 2s - 4pi*(8-2s) = 0
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2s - 4pi*(8-2s) = 0
s = 16pi - 4pi*s
s*(4pi + 1) = 16pi
s = 16pi/(4pi + 1)
s =~ 3.705 inches
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dA/ds = 2s - 4pi*(8-2s)
2nd derivative = 2 + 8pi which is positive --> a minimum
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You can do it using dA/dC in a similar manner.
I prefer making the side of the square the variable.
You can put this solution on YOUR website! Let L = the length of the wire
Let s = the perimeter of the square, then s/4 = the length of one side
Then the circumference of the circle, C = L - s
Area of the square = (s/4)^2
Area of the circle = C^2/(4*pi)
A = the sum of the areas = (s/4)^2 + C^2/(4*pi)
In terms of C, since s = L - C, we can write
A = (L-C)^2/16 + C^2/(4*pi)
For the sum of the areas to be a minimum, dA/dC = 0:
0 = 2(L-C)(-1)/16 + 2C/(4*pi)
0 = (C-L)/8 + C/(2*pi)
Solve for C in terms of L:
C(1/8 + 1/(2*pi)) = L/8
C(1 + 4/pi) = L -> C = L/(1 + 4/pi)
Substituting L = 16 gives C = 7.04 inches
Therefore, the perimeter of the square is s = 16 - 7.04 = 8.96 inches
