SOLUTION: The monthly revenue achieved by selling x boxes of candy is calculated to be $ x(5-0.05x). The wholesale cost of each box of candy is $1.50. How many boxes must be sold each mont

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: The monthly revenue achieved by selling x boxes of candy is calculated to be $ x(5-0.05x). The wholesale cost of each box of candy is $1.50. How many boxes must be sold each mont      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 39412: The monthly revenue achieved by selling x boxes of candy is calculated to be
$ x(5-0.05x). The wholesale cost of each box of candy is $1.50. How many boxes must be sold each month to maximize profit?
What is the maximum profit?
(Revenue=Cost+Profit)

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
This is a workout for my calculator!
It's asking for Profit, so I rewrite
Revenue = Cost + Profit
Profit = Revenue - Cost (P = R - C)
R+=+x%285+-+.05x%29
R+=+5x+-+%285%2F100%29x%5E2
The cost of 1 box is 1.50, so the cost of x boxes is 1.5x
P+=+R+-+C
P+=+5x+-+%285%2F100%29x%5E2+-+1.5x
P+=+-%281%2F20%29x%5E2+%2B+3.5x
If I find the roots, the maximum will be midway between
the 2 roots. Set P = 0 to find roots.
-%281%2F20%29x%5E2+%2B+3.5x+=+0
multiply both sides by -1
%281%2F20%29x%5E2+-++3.5x+=+0
x%5E2+-+70x+=+0
x%28x+-+70%29+=+0
The roots are 0,70. The midpoint is (0 + 70)/2 = 35
I want to test the equation for P to see if 35 is really
a maximum. I will find P(34), P(35), and P(36).
P(34) and P(36) should both be slightly less than P(35)
P+=+-%281%2F20%29x%5E2+%2B+3.5x
P%2834%29+=+-%281%2F20%2934%5E2+%2B+3.5%2A34
P%2834%29+=+-57.8+%2B+119
P%2834%29+=+61.2
-------------------------------
P%2836%29+=+-%281%2F20%2936%5E2+%2B+3.5%2A36
P%2836%29+=+-64.8+%2B+126
P%2836%29+=+61.2
-------------------------------
P%2835%29+=+-%281%2F20%2935%5E2+%2B+3.5%2A35
P%2835%29+=+-61.25+%2B+122.5+=+61.25
So, 35 boxes must be sold each month to maximize Profit,
which is $61.25