SOLUTION: 1. A company that produces cells for solar collectors finds through experience that on the average it can sell x cells per day at a selling price (in dollars) of p(x) = 100 – 0.05x

Algebra ->  Functions -> SOLUTION: 1. A company that produces cells for solar collectors finds through experience that on the average it can sell x cells per day at a selling price (in dollars) of p(x) = 100 – 0.05x      Log On


   



Question 129106: 1. A company that produces cells for solar collectors finds through experience that on the average it can sell x cells per day at a selling price (in dollars) of p(x) = 100 – 0.05x (p is called the demand function). The cost function C(x) for producing x cells per day is C(x)= 4000 + 60x – 0.01x^2

A. Write the revenue function R(x).

B. Write the profit function P(x).


C. Find the value of x that maximizes profit, and find the maximum profit.

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
a)

The revenue can be found by this general formula:

Revenue=Number of units sold*price of each unit



So in our case, R%28x%29=x%2AP%28x%29


R%28x%29=x%2AP%28x%29 Start with the given equation


R%28x%29=x%2A%28100-0.05x%29 Plug in P%28x%29+=+100+%96+0.05x


R%28x%29=100x-0.05x%5E2 Distribute


R%28x%29=-0.05x%5E2%2B100x+ Rearrange the terms


So the revenue function is

R%28x%29=-0.05x%5E2%2B100x+






b)

The profit can be found by:

Profit=Revenue-Cost

So

P%28x%29=R%28x%29-C%28x%29


P%28x%29=%28-0.05x%5E2%2B100x%29-%284000+%2B+60x-0.01x%5E2%29 Plug in R%28x%29=-0.05x%5E2%2B100x+ and C%28x%29=4000+%2B+60x+%96+0.01x%5E2


P%28x%29=-0.05x%5E2%2B100x-4000-60x+%2B+0.01x%5E2 Distribute the negative


P%28x%29=-0.04x%5E2%2B40x-4000 Combine like terms






To find the max profit, we need to graph the profit function P%28x%29=-0.04x%5E2%2B40x-4000


+graph%28+500%2C+500%2C-740%2C1260%2C-10523%2C7649%2C-0.04x%5E2%2B40x-4000%29


And by using the calculators vertex function, we find that the vertex occurs at x=500


Now plug in x=500 into P%28x%29=-0.04x%5E2%2B40x-4000


P%28x%29=-0.04%28500%29%5E2%2B40%28500%29-4000

P%28x%29=6000 Simplify


So the maximum profit is $6,000