Linear Demand Function

[mamath]

Can somebody explain how to answer this?

if linear demand function yields demand of 10000 units at a price of $1.00, and demand is 6000 units at a price of $1.20, what is the maximum revenue that can be generated?

 

[JeffM]

Can somebody explain how to answer this?

if linear demand function yields demand of 10000 units at a price of $1.00, and demand is 6000 units at a price of $1.20, what is the maximum revenue that can be generated?

I do not have a lot of time so if this response is cryptic, let me know, and I'll be back this evening to expand.

Let S = units sold
P = price per unit
R = revenue.

Obviously R = P * S.
If you have S as a function of P, then you can find the value of P that maximizes R by setting R's derivative = 0, right?
But you CAN calculate S as a function of price.
You are told that the function S= f(P) is linear, and you are given two points on that line. So you can determine the equation of that line through algebra.

Make sense?

 

[mamath]

that helps a bit, but i am struggling to apply it to this practice problem successfully.

can you (or anybody else) help walk me through this problem so i can see the step-by-step process for determining the solution?

many thanks!

 

[Deleted member 4993]

Can somebody explain how to answer this?

if linear demand function yields demand of 10000 units at a price of $1.00, and demand is 6000 units at a price of $1.20, what is the maximum revenue that can be generated?

I'll do a similar but different problem:

if linear demand function yields demand of 20000 units at a price of $5.00, and demand is 11000 units at a price of $14.00, what is the maximum revenue that can be generated?

First we need to establish price-demand equation. (d and p in thousands)

(14 -p)/(14-5) = (11 - d)/(11-20)

p = 25 - d

Then

R(evenue) = d * p = d(25-d)

This is a parabola (upside down) with vertex at d = 25/2 (maximum R)

R[sub:2ggco9bq]max[/sub:2ggco9bq] = 25/2 (25 - 25/2) = 625/4 = 156.25

Maximum revenue = $156,250

 

[JeffM]

I'll do a similar but different problem: (I am going to elaborate a bit on SK's excellent example.)

if linear demand function yields demand of 20000 units at a price of $5.00, and demand is 11000 units at a price of $14.00, what is the maximum revenue that can be generated?

First we need to establish price-demand equation. (d and p in thousands) SK meant to say d in thousands.

(14 -p)/(14-5) = (11 - d)/(11-20) He uses the information given about the demand function to develop a functional relationship between quantity demanded and sales price. That information is that the function is linear plus two points on the line are given.

p = 25 - d It is standard in modern economics to treat demand as the dependent variable and price as the independent variable so most economists would write this as d = 25 - p. This makes intuitive sense because d falls as p goes up. Of course both expressions lead to the same arithmetic results.

Then

R(evenue) = d * p = d(25-d)

This is a parabola (upside down) with vertex at d = 25/2 (maximum R)

R[sub:3k6xvs9u]max[/sub:3k6xvs9u] = 25/2 (25 - 25/2) = 625/4 = 156.25

Maximum revenue = $156,250

An alternative and more general method once demand is specified as a function of price is to use calculus.
R = d * p = (25 - p) * p = 25p - p[sup:3k6xvs9u]2[/sup:3k6xvs9u]. You have a a revenue function in terms of p, which is the standard way that economists now show it.
To maximize R, we find where the derivative equals zero.
dR/dp = 25 - 2p.
dR/dp = 0 when p = 12.5.
When p = 12.5, R = 25p - p[sup:3k6xvs9u]2[/sup:3k6xvs9u] = (25 * 12.5) - 12.5[sup:3k6xvs9u]2[/sup:3k6xvs9u] = 12.5(25 -12.5) = 12.5[sup:3k6xvs9u]2[/sup:3k6xvs9u] = 156.25.
Maximum revenue = $156,250. This is the same as SK's answer, but it uses the method most common in economics since the time of Jevons.