Answer to Question #1918 in Calculus for vigi
(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?
S(sq) = a2, S(tr) = √3/4 b2
b = (25 - 4a)/3
Thus the total square is
S(tot) = S(sq) + S(tr) = a2 + √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
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