revenue of a product?

Question

the number of units (n) of a certain commodity sold at price p is given by n = -0.2p + 50. The revenue is the product of the selling price and the number of units sold. Find the maximum revenue for this commodity.

the answer is $3,125, but i need to know how to get there

Answer 1

The revenue formula is r=np. We have a relationship of n and p, so we can substitute the function n(p) into n in the r formula.

r=(-0.2p + 50)p = -0.2p^2 + 50p.

Now to find the price that results in the maximum revenue, take the derivative, set it equal to 0 and solve for p:

r' = -.4p + 50 = 0
p=125

Plug this into the r formula to find the maximum revenue:
r=-0.2p^2 + 50p=-0.2(125)^2 + 50(125) = $3125

Answer 2

Maximum revenue = $3125. Here's how to obtain this:

Revenue R (in $)### = np = [50 - 0.2 p] p = 50 p - 0.2 p^2 ......(A)

The maximum revenue can be found in at least two ways:

1. By calculus. From equation (A), R is at an extremum when dR/dp = 0, i.e. when 50 - 0.4 p = 0, or p = 50 / 0.4 = 125.

At this point, R = [50 - 0.2 p] p = 25 x 125 = 3125. Since

[d^2]R/dp^2 = - 0.4, and that is NEGATIVE, this is a MAXIMUM.

2. By algebra alone. From equation (A),

R = (1/5)[250 p - p^2] = (1/5) [(125)^2 - (p - 125)^2]. ......(B)

Since a perfect square term is being subtracted in equation (B), the MAXIMUM occurs when that subtracted square term is ZERO, that is when p = 125. And, at that point,

R = (1/5) (125)^2 = 3125. QED

Live long and prosper.

### You didn't GIVE any currency unit originally, but since you later made it clear you were expecting the answert in dollars, I have now inserted that currency unit. It makes no difference to the calculations, of course.

Answer 3

Max revenue = 3125 with n = 25

Answer 4

R=np=(25)(125)=3125

R=np=(-.2p+50)p=-.2p^2+50p

dR/dp= -.4p+50=0, p=125

n=-.2p+50= -.2(125)+50=25

R=np=(25)(125)= $3,125