Awire is to be cut into two pieces. one piece will be bent into a square, and the other piece will be bent into a circle. if the total area enclosed by the two pieces is to be 64 cm2,
what is the minimum length of wire that can be used?
what is the maximum length of wire that can be used?

Step-by-step explanation:

The relationships between the radius of a circle and its circumference and area are ...

  C = 2πr

  A = πr²

The relationships between the side length of a square and its perimeter and area are ...

  P = 4s

  A = s²

So, the length of wire will be ...

  w = C + P

  w = 2πr + 4s

subject to the constraint that the sum of areas is 64 cm²:

  πr² + s² = 64

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Using the method of Lagrange multipliers to find the extremes of wire length, we want to set the partial derivatives of the Lagrangian (L) to zero.

  L = 2πr + 4s + λ(πr² +s² -64)

  ∂L/∂r = 0 = 2π +2πλr . . . . . . [eq1]

  ∂L/∂s = 0 = 4 +2λs . . . . . . . . [eq2]

  ∂L/∂λ = 0 = πr² +s² -64 . . . . [eq3]

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Solving for λ, we find ...

  0 = 1 +λr . . . . divide [eq1 by 2π

  λ = -1/r . . . . . . subtract 1, divide by r

Substituting into [eq2], we get ...

  0 = 4 + 2(-1/r)s

  s/r = 2 . . . . . . . . . .add 2s/r and divide by 2

This tells us the maximum wire length is that which makes the circle diameter equal to the side of the square.

Substituting the relation s=2r into the area constraint, we find ...

  πr² +(2r)² = 64

  r = √(64/(π+4)) = 8/√(π+4) ≈ 2.99359 . . . . cm

and the maximum wire length is ...

  2πr +4(2r) = 2r(4+π) = 16√(4+π) ≈ 42.758 . . . cm

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The minimum wire length will be required when the entire area is enclosed by the circle. In that case, ...

  πr² = 64

  r = √(64/π)

  C = 2πr = 2π√(64/π) = 16√π ≈ 28.359 . . . cm

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Comment on the solution method

The method of Lagrange multipliers is not needed to solve this problem. The alternative is to write the length expression in terms of one of the figure dimensions, then differentiate with respect to that:

  w = 2πr + 4√(64-πr²)

  dw/dr = 2π -4πr/√(64-πr²) = 0

  64 -πr² = 4r²

  r = √(64/(π+4)) . . . . same as above

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Comment on the graph

The attached graph shows the relationship between perimeter and circumference for a constant area. The green curve shows the sum of perimeter and circumference, the wire length. The points marked are the ones at the minimum and maximum wire length.

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