SOLUTION: A company that manufactures flash drives knows that the number of drives x it can sell each week is related to the price p , in dollars, of each drive by the equation x=12

Question 1183538: A company that manufactures flash drives knows that the number of drives
x
it can sell each week is related to the price
p
, in dollars, of each drive by the equation
x=1200-100p
Find the price
p
that will bring in the maximum revenue. Remember, revenue (R) is the product of price (p) and items sold (x), in other words, R=xp

Found 3 solutions by ikleyn, ankor@dixie-net.com, Solver92311:
Answer by ikleyn(52865)   (Show Source): You can put this solution on YOUR website!
.
A company that manufactures flash drives knows that the number of drives x
it can sell each week is related to the price p, in dollars, of each drive by the equation
x=1200-100p
Find the price p that will bring in the maximum revenue.
Remember, revenue (R) is the product of price (p) and items sold (x), in other words, R=xp
~~~~~~~~~~~~~~~

The revenue is this function of p (of the price) R(p) = p*x = p*(1200-100p) It is a quadratic function of p with the zeroes p = 0 and p = 1200/100 = 12. The maximum of this quadratic function (its vertex) is halfway between the roots at x= 6. So, the optimal price is p= 6. The problem is just solved.

----------------

My common sense tells me that the problem's description is DEFECTIVE and must be corrected.

I leave making this correction in your care.

Answer by ankor@dixie-net.com(22740)   (Show Source): You can put this solution on YOUR website!
A company that manufactures flash drives knows that the number of drives x it can sell each week is related to the price p, in dollars, of each drive by the equation x=1200-100p
Find the price p that will bring in the maximum revenue.
:
R = p * x
replace x with 1200-100p
R(p) = p(1200-100p)
R(p) = 1200p - 100p^2
or
R(p) = -100p^2 + 1200p
using the vertex formula: p = -b/(2a) where a = -100, b = 1200
p =
p = $6 is the price that will give max revenue

Answer by Solver92311(821)   (Show Source): You can put this solution on YOUR website!

is an extreme point

Therefore is a maximum.

John

My calculator said it, I believe it, that settles it

From
I > Ø

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