maximum revenue due tonite please help!!!

[yunvme09]

1 pt) The revenue function in terms of the number of units sold ,x, is given as
R=390x- 0.3x^2

where R is the total revenue in dollars. Find the number of units sold x that produces a maximum revenue?

 

[arthur ohlsten]

R=-.3x^2 +390x
R=-.3[x^2-1300x] complete the square
R=-.3 [ x^2 -1300x + (1300/2)^2] +[.3][1300/2]^2
R=-.3[x-1300/2]^2 + (.3)(650^2) a parabola open down, vertex at 650,[(.3)(650^2) ]

maximum R = 650^2(.3)
Rmax=650 [ 195] answer
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another way

R=-.3x^2+390x take derivative and set =0
dR/dx= -.6x +390
0=-.6x+390
.6x=390

arthur
x=650

maximum R= -.3[650^2] +390[650]
R max= 650[390-195]
R max= 650[195]