Math Question About Finding Revenue?
The John Deere Company has found that the revenue from sales of heavy-duty tractors is a function of the unit price p (in dollars) that it charges. If the revenue R is
R(p)= -1/2p*2+1900p
What unit price p (in dollars) should be charged to maximize revenue? What is the maximum revenue?
PS: (*) means exponent
October 22nd, 2009 at 5:44 pm
Do you know about derivatives?
Find the derivative of -1/2p*2 + 1900p and set it equal to zero. Zero is actually the slope of the line that crosses R(p) at its maximum in the graph.
The derivative is -p + 1900 = 0. Solve for p. p=1900 which will maximize revenue.
October 22nd, 2009 at 11:29 pm
You have to take the derivative of the Revenue function and equal it to zero
R(p)=-1/2p^2+1900p
R’(p)=-p+1900=0
p=1900
The maximum Revenue is:
R(1900)=-1/2(1900)^2 + 1900(1900)
R(1900)=-1/2(1900)^2 + (1900)^2
R(1900)=1/2(1900)^2
R(1900)=1,805,000
I hope this can be useuful
David
October 23rd, 2009 at 5:00 am
R’=-p+1900
p=1900
October 23rd, 2009 at 5:51 am
Take the derivative of the price function, set it to zero and solve for p