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September 1, 2014

September 1, 2014

Posted by **susey** on Monday, March 12, 2007 at 10:25pm.

This looks like a problem in 'linear programming'

let s represent the number of suits and d the number of dresses.

then 2s + d <= 100 and 3s + 2d <= 180

a d vs

graph these two inequations with s and d as the axes forming a polygon with the s and d axes.

Now consider the equation Profit = 108s + 60d

Shift this equation away from the origin into the first quadrant, until the line p=108s+60d reaches the farthest vertex from the origin of your polygon.

Your polygon only has one vertex in the interior region of the first quadrant, I think it is s=20 and d=30 for a maximum profit of $3960

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