Question 965267: Must develop a quadratic function in order to solve the problem: A store sells 6000 gallons of paint each week for $15 per gallon. For each $1 decrease in the price, the store will sell an additional 1000 gallons each week. What should the sale price be to maximize the income?
Found 2 solutions by Edwin McCravy, josgarithmetic:
Answer by Edwin McCravy(20054)   (Show Source):
You can put this solution on YOUR website! Must develop a quadratic function in order to solve the problem: A store sells
6000 gallons of paint each week for $15 per gallon. For each $1 decrease in the
price, the store will sell an additional 1000 gallons each week. What should the
sale price be to maximize the income?
Let x = the number of dollar decreases.
Let y = the income with x dollars degreases.
For each $1 decrease in the price, the store will sell an additional 1000 gallons each week.
Number of gallons sold after x dollar decreases = 6000 + 1000x
Price per gallon after x dollar decreases = $15-$x
Income = y = (6000+1000x)(15-x)
y = 90000-6000x+15000x-1000x^2
y = 90000+9000x-1000x^2
y = -1000x^2+9000x+90000
Compare to
y = ax^2+bx+c
a= -1000, b = 9000
Use vertex formula:
x-coordinate of vertex = -b/(2a) = -(9000)/(2*-1000) = -9000/(-2000) = 4.5
Therefore decreasing the price 4.5 dollar decreases, which means a reduction
of $4.50 will yield the maximum income.
So the sale price should be $15-$4.50 = $10.50 per gallon, to maximixe
income.
Edwin
Answer by josgarithmetic(39616)  (Show Source):
You can put this solution on YOUR website! PRICE*UNITSSOLD=REVENUE
15*6000=R
(15-1)*(6000+1000)=R(1)
(15-2)(6000+2*1000)=R(2)
(15-3)(6000+3*1000)=R(3)
.
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The variable here is the amount of decrease from the initial 15 dollar per gallon price, so the variable choice this way can be x, the amount of price DECREASED off the $15.
Formula for revenue is the function  .
Do the simplifying into general form, and you'll have the quadratic equation to which you can apply the general solution for a quadratic equation; but what you want is the value for x in the MIDDLE of the zeros of R. That will be the maximum, because R(x) is a function with a maximum point (according to the coefficient on x^2 being negative).
If that analysis makes sense, after you read and think about it carefully, then you know what to do. Just remember, as analyzed here, the price you are looking for is 15 MINUS the middle value between the zeros.
Note,  , so when you examine the roots,
Dividing both sides by 1000,
...and again, look for the x value exactly between the two zeros.
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