Raymond B. answered 05/20/22
Math, microeconomics or criminal justice
Profit = P = R-C
P = -x^2/2 + 85x - (45x + 300)
P = -x^2/2 +85x-45x -300
P = -x^2/2 + 40x -300
P = 50 = -x^2/2 +40x -300
x^2/2 -40x +350 = 0
x = 40/2(1/2) + or - sqr(40^2 -4(350))
x = 40 + or - sqr(200)
x = 40 + or - 10sqr2
x = about 40+ or -14.14) solutions are: (40-10sqr2, 40+10sqr2)
or with a semi-colon, then (20-10sqr2; 20+10sqr2)
x = about 25.86 or 54.14
x = about each make a profit of 50
solutions are (25.86, 54.14)
or if you really want a semi-colon, ;, which is strange, then (25,86; 54.14)
Maximum P is when P' = 0
P' = -x + 40 = 0
x = 40
P(40) = -(40^2)/2 + 40(40) - 300 = -800+1600 - 300 = 500<2500
A profit of 2500 is Not possible, on the information given.
But the Profit function is profit per some time period, such as maybe 1 month
then you could earn a profit of 2500 for a time period of 5 months, 500x5
You might graph the Profit function. It's a downward opening parabola
with vertex = maximum profit point = V(40, 500)