set R(x) = C(x)
x(50-1.25x) = 160 + 10x
50x - 1.25x^2 -160 - 10x = 0
1.25x^2 -40x + 160 = 0
x^2 - 32x + 128 = 0
complete the square
x^2 - 32x = -128
x^2 - 32x + 256 = -128+256
(x-16)^2 = 128
x-16 = ±√128
x = 16± √128 = 4.686 or 27.314 , but 0 < x < 20
so x = 4.686
I will assume your R, C, and P represent
Revenue, Cost and Profit respectively, thus
P(x) = x(50-1/25x) - (160 + 10x)
= 50x - 1.25x^2 - 160 - 10x
P'(x) = 50 - 2.5x - 10
= 0 for a max of P
2.5x = 60
x = 24
max profit = P(24) = ....
I will let you push them buttons.
The research department in a company that manufactures AM/FM clock radios established the following cost, and revenue functions:
R(x) = x(50-1.25x)
C(x) = 160 + 10x
where 0 < x < 20.
a) Determine when R = C (to the nearest thousand units. e.g. 6.247)
Quantity x = ________ thousand units.
b) Determine the maximum profit (to the nearest thousand dollars).
Profit P = _______ thosand dollars.
My answers don't seem to be correct (x = 16 and P = 160) Please help!
3 answers
Is max profit 80 than? I am plugging it back into P(x) = = 50x - 1.25x^2 - 160 - 10x right?
yes!, good job
1 answer