Answered

The John Deere company has found that the revenue from sales of heavy-duty tractors is a function of the unit price p it charges. If the revenue R is:

R=-1/2p^2+1900p

What unit p should be charged to maximize revenue?
What is the maximum revenue?

Answer :

nandhini123

Answer:

1900 unit p should be charged to maximise revenue and 1805000  is the maximum revenue

Step-by-step explanation:

The equation represents a parabola that is looking down and has the maximum, we have to find the vertex of equation.(p,R)

vertex = [tex]-\frac{b}{2a}[/tex]  

Here b = 1900

a =[tex](-\frac{1}{2})[/tex]

On substituting the values

[tex]-\frac{1900}{2(-\frac{1}{2})} = 1900[/tex]   it is your x for vertex

put it in your equation and calculate:

[tex]R= -\frac{1}{2}(1900)^2 +1900 \times 1900[/tex] = -1805000 + 3610000= 1805000

price must be equal 1900 to maximise the revenue = 1805000

Other Questions