Question #1918

A piece of wire 25 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. (Give your answers correct to two decimal places.)
(a) How much wire should be used for the square in order to maximize the total area?
(b) How much wire should be used for the square in order to minimize the total area?

Expert's answer

Denote the side of the square as a, and the side of equilateral triangle as b.

S(sq) = a^{2}, S(tr) = √3/4 b^{2}

b = (25 - 4a)/3

Thus the total square is

S(tot) = S(sq) + S(tr) = a^{2} + √3/36 (25 - 4a)^{2}

Let's find the derivative dS/da and assume it equal to zero:

2a - 2√3/9 (25 - 4a) = 0

a (1 + √3/9) = 25√3/9

a = 4.03 - this point is local minimum of the fucntion S(a). - this side of the square should be used to minimize of the total area.

The maximum possible value of a is 25/4, minimum - 0

S(a = 6.25) = 39.063

S(a = 0) = 30.07

Thus the maximum area will be abtained at a = 6.25

S(sq) = a

b = (25 - 4a)/3

Thus the total square is

S(tot) = S(sq) + S(tr) = a

Let's find the derivative dS/da and assume it equal to zero:

2a - 2√3/9 (25 - 4a) = 0

a (1 + √3/9) = 25√3/9

a = 4.03 - this point is local minimum of the fucntion S(a). - this side of the square should be used to minimize of the total area.

The maximum possible value of a is 25/4, minimum - 0

S(a = 6.25) = 39.063

S(a = 0) = 30.07

Thus the maximum area will be abtained at a = 6.25

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