Answer:
a) P(c) = xc² + yc + z, and P(.15) = $124, P(.05) = 0, P(.17) = 0. Thus:
(#1) .0225x + .15y + z = 124
(#2) .0025x + .05y + z = 0
(#3) .0289x + .17y + z = 0
Subtracting #2 from #3: (#4) .0264x + .12y = 0
Subtracting #2 from #3: (#5) .02x + .1y = 124
Subtracting 1.2×#5 from #4: .0024x = -148.8 → x= -62000
Using this value of x in #5: -1240 + .1y = 124 → y = 13640
Using x and y in #1: -1395 + 2046 + z = 124 → z = -527
P(c) = -62000c² + 13640c - 527
b) To find the maximum of P(c), find a such that P'(a) = 0
-124000c + 13640 = 0
c = .11
At this price, his profit would be $223.20
