A company's revenue from selling x units of a new software game is R(x)=1000x-20x^2 dollars. if sales are increasing at a rate of 50 units per day, find how fast the revenue is growing at a time when 20 units have been sold.
Do you know calculus?
You need the RATE of revenue growth.....this is the SLOPE of the revenue function at x = 20
The DERIVATIVE of the revenue funtion will give you the slope
the derivative is R'(x) = 1000 - 40x at x = 20 this equals 200 dollars/unit (edited.....had wrong units )
200 dollars/unit x 50 units/day = 10000/day thank you
I have been studying Calculus for 13 YEARS! this is a question that is far beyond my years of experience. Do i know calculus?? YES i have 4 masters of Calculus, the answer is 10k per day by the way, thankyou very much.
Guest was just trying to be helpful. No need to be snarly about it.
I had the same thoughts as them to start with.
\(R=1000x-20x^2\\ \frac{dx}{dt}=50\\ Find \;\;\frac{dR}{dt}\;\;when\;\;x=20\\~\\ \frac{dR}{dt}=\frac{dR}{dx}*\frac{dx}{dt}\\ \frac{dR}{dt}=(1000-40x)*50\\ \text{When x=20}\\ \frac{dR}{dt}=(1000-40*20)*50\\ \frac{dR}{dt}=(1000-800)*50\\ \frac{dR}{dt}=10 \;000\\ \)
That is a daily rate.
Revenue= R = 1000x - 20x^2
R = 1000x - 20x^2
dR/ dT = dR / dx * dx / dT
So
dR / dT = (1000 - 40x ) * [ 50 units per day ]
And when x = 20
dR /dT = [1000 - 40 (20)] * [ 50]
dR /dT = [ 1000 - 800 ] * [50]
dR /dT [ 200] * [50 ] = 10,000
The revenue is changing by $10,000 when 20 units are sold