
John K. answered 08/16/15
Tutor
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Math and Engineering Tutor, Professional Engineer
The problem is not concisely given. It seems to assume 500 students will attend which makes the problem ill defined. If we assume the number of students can vary then let x be the number of 0.20 increments in cost and the number of 25 student increments then the cost is (3+.2x)(500-25x) or -5x^2+25x+1500. We can differentiate to find the maximum or -10x+25=0 or the number of 0.2 increments is 2.5. and the cost is 3.50 and the number of students is 500-62.5=437.5 and the maximum profit is 437.5*3.5 = 1531.25.


Christopher G.
Err, +2.5 (obviously)
Report
08/29/15
Christopher G.
x =-b/2a
= -25/(-10)
08/29/15