SOLUTION: 3) The profit function for Wannamaker Trophies is {{{P(x) = -0.4x^2 + fx - m}}}, where f represents the design fee for a customer’s awards and m represents the monthly office rent

Algebra ->  Graphs -> SOLUTION: 3) The profit function for Wannamaker Trophies is {{{P(x) = -0.4x^2 + fx - m}}}, where f represents the design fee for a customer’s awards and m represents the monthly office rent      Log On


   

Question 798620: 3) The profit function for Wannamaker Trophies is P%28x%29+=+-0.4x%5E2+%2B+fx+-+m, where f represents the design fee for a customer’s awards and m represents the monthly office rent. Also, P represents the monthly profit in dollars of the small business where x is the number of awards designed in that month.
a) If $80 is charged for a design fee, and the monthly studio rent is $1,600; write an equation for the profit, P, in terms of x.
b) How much is the profit when 50 award designs are sold in a month? Answer:
c) How many award designs must be sold in order to maximize the profit? Show your work algebraically.
d) What is the maximum profit?
Answer by Finavon(81) About Me  (Show Source):
You can put this solution on YOUR website!
a) P%28x%29+=+-0.4x%5E2+%2B+80x+-+1600+

b) x=50
P=+-0.4%2A2500+%2B+4000+-+1600+=+1400

c)
Graph function takes 3 pairs of numbers before the equation.
First pair define the size of the image on the screen (width then height).
Set this to see the graph on the screen.
Second pair are the range of values on the x axis (min then max)
Third pair are the range of values on the y axis (min then max)
You can calculate some vaues, but I just set these wide to see the function. Using (-1000, 1000) for both:
graph%28300%2C200%2C-1000%2C1000%2C-1000%2C1000%2C80x-0.4x%5E2-1600%29
Inspecting the result shows that the region of interest, (in this case the values where the maximum value of y occurs) lies at higher values of y (try -500,5000). The x values can be narrowed down (try -100,400).
graph%28300%2C200%2C-100%2C400%2C-500%2C5000%2C80x-0.4x%5E2-1600%29
Further adjustment (to -50,200,-300,2700) produces
graph%28300%2C200%2C-50%2C200%2C-300%2C2700%2C80x-0.4x%5E2-1600%29
Inspecting this shows the max lies around 100 sales
d)
Max Profit @ x=100 sales
P=-4000+8000-1600=$2400
That sale brought in $80 and cost 199/2.5 =$79.60
Next sale brings in $80 but costs 201/2.5=$80.40