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# The "aha!" problems - 3 minimum areas

## Question

A while back I started a series of puzzles that seem difficult but

have not-so-difficult answers. I can't find any of them, so I don't

know how many exist. There must have been at least two, so I'll

number this one 3. Have fun.

A monotonic increasing function f (x) is cut above and below by horizontal lines that intersect it at f (x1) and f (x2).

A vertical line is drawn through a point on the curve (black dot) between x1 and x2.

The curve and the lines define the green and red areas shown on the figure.

You are to find the point on the curve that minimizes the sum of these areas.

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## 2 answers to this question

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Draw a line parallel to the 2 horizontal lines and midway between them. Let it cut the curve at O=f(x0). Notice that the curve is always below the middle line, to the left of O and always above it, to the right of O (monotonic increase)
Assume the big dot is at the left of O at point f(x1). if it moves to the right to point f(x2) (x2<x0) its clear that area gained is smaller than the area lost as segment x1-x2 is below the middle line. Hence we can keep moving to the right till we reach x0 as the total area will keep reducing. After point x0, if we keep moving to the right, the situation will reserve.
Hence the big dot lies where a line equidistant and parallel to the 2 horizontal lines cuts f(x).
ALTERNATIVE APPROACH:
to find minimum area S = Int(f(x))x1->x + ( (x2-x)*( f(x2) - f(x1) ) - ( Int(f(x))x->x2) ) where Int(f(x))a->b denotes integral of f(x) from x=a to x=b. (How does 1 write integrals in here?)
Differentiating S wrt x and equating to 0 results in f(x) = (f(x1)+f(x2))/2
Same as in the first approach

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Nice job m00li.

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