Arthur D. answered 11/01/17
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How many $0.15 increases will maximize the profit ?
($0.95+$0.15x)(10,000-500x)=0 where x is the number of $0.15 increases
9500+1500x-475x-75x2=0
-75x2+1025x+9500=0
the maximum occurs at the vertex, (h,k), of the parabola
x=-b/2a
x=-1025/(-150)
x=6.8333...
seeing that you have to have whole number increases, you need 7 $0.15 increases
$0.95+7*$0.15=$0.95+$1.05=$2.00
$2.00*(10,000-7*500)
$2.00*(10,000-3500)
$2.00*6500=$13,000 is the maximum revenue
if there were 6 increases...
$1.85*7000=$12,950
if there were 8 increases...
$2.15*6000=$12,900
$2.00 is the price of the bolt and the maximum revenue is $13,000 (7 increases of $0.15)