A small company manufactures a total of x products per week. The production cost is modelled by the function y=50 + 3x. The revenue is given by the function 6x - x^2/100. How many items per week should be manufactured to maximize the profit for the company?
Please help! The answer is 150 but I don't know how to come to it.
Cost: c(x)=50+3x
Revenue: r(x)=6x-x^2/100
Profit=Revenue - Cost
Therefore:
p(x)=6x-x^2/100 - 50+3x
p(x)=-x^2/100-3x-50
(-b/2a, y)
(-3/2(1/100)
(150, y)
parabola, complete square
-x^2/100 + 3x - 50 = p
x^2 - 300 x + 5000 = -100 p
x^2 - 300 x = - 100 p - 5000
x^2 - 600 x + (150)^2 = -100p+22,500
(x-150)^2 = -100 (p-225)
vertex at x = 150, p = 225
so once again, 150
P(x)=6x-x^2/100 - 50+3x
P(x)=-x^2/100 + 3x-50
a=-x^2/100
b=3x
c=50
h= -b/2a
h= -3/2(1^2/100)
h= -3/0.02
h= -150
aka positive 150 when graphed.
profit = 6x - x^2/100 - 50 - 3x
p = -x^2/100 + 3 x - 50
dp/dx = 0 at max = -x/50 + 3
x = 50 * 3 = 150
jgmhmbku,o
profit = 6x - x^2/100 - 50 - 3x
p = -x^2/100 + 6 x - 50
dp/dx = 0 at max = -x/50 + 6
x = 50 * 6 = 300
whooops, you have not have derivatives
parabola, complete square
-x^2/100 + 6x - 50 = p
x^2 - 600 x + 5000 = -100 p
x^2 - 600 x = - 100 p - 5000
x^2 - 600 x + (300)^2 = -100p+85,000
(x-300)^2 = -100 (p-850)
vertex at x = 300, p = 850
so once again, 300