Posted by Ashley on Tuesday, January 29, 2013 at 11:52am.
Basically you want the smallest area bounded by the x- and y-axes, and any line passing through (65,55).
A line through (65,55) with slope m will have intercepts at y = 55-65m and x=65+55/m
So, the area of the enclosed triangle is
a(m) = 1/2 (55+65m)(65+55/m)
a(m) = 25/2m (11-13m)^2
da/dm = 25/2 (121/m^2 - 169)
da/dm=0 when m = ±11/13
We know we need a negative slope, so m = -11/13
The minimum area is thus 7150
oops on typos
That should be 1/2 (55-65m)(65-55/m)
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