Asked by Anonymous
Twenty feet wire is used to make two figures? What is the maximum areas of enclosed figures.
Answers
Answered by
Steve
max area for a given perimeter is a circle. Is the wire evenly divided?
If so, c = p/2, so r = p/4π, and the total area is
a = 2(πr^2) = 2π(p^2/16π^2) = p^2/8π
If not, then if we have a tiny circle, (area effectively zero), then r = p/2π and
a = πr^2 = πp^2/4π^2 = p^2/4π
I suspect there is something missing here from the problem statement.
If so, c = p/2, so r = p/4π, and the total area is
a = 2(πr^2) = 2π(p^2/16π^2) = p^2/8π
If not, then if we have a tiny circle, (area effectively zero), then r = p/2π and
a = πr^2 = πp^2/4π^2 = p^2/4π
I suspect there is something missing here from the problem statement.
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