Econ 407:

More on differentitation. The graph of y= 1/x and its derivative are shown below. Note that the green curve shows the derivative of the function at every point. The numerical value of the derivative (slope) can be read off the Y axis.

> plot(1/x,x=1..5);

> dy:=diff(1/x,x);

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dy := -1/x^2

> plot({1/x,dy},x=1..5);

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Let us apply the concept of derivative to find the maximum point on the total revenue function. The Total revenue function: Tr is found by multiplying price times quantity.

> qd:=50-4*p;

> Tr:=p*(50.0-4*p);

Assume this firm is a monoploy; like the local Cable Company; it can charge anything it please. Question:

What price will maximize their total revenue.? First plot the total revenue curve, and then solve for the maximum point by setting the first derivative of the TR function equal to zero.

> plot(Tr, p=0..10);

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More formally, to find the price that maximizes TR, we should take first derivative of TR with respect to p, and set equal to zero

solve for p as shown below:

> Mr:=diff(Tr, p);

> solve(50.0-8*p=0, p);

> eval(qd,p=6.25);

The firm will produce 25 units, and charge 6.25 to mamimize profits.

What if the firm is operating under competition. In this case the firm must adjust quantity and accept the price as a given. We need to reqrite the demand and TR functions in terms of q.

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> restart;

Now let us solve the demand function in terms of price: which will be p= -0.25qd+ 12.5.

This can be done using the following comannd.

> solve(qd-50.0+4*p,p);

> p:=-.25*qd+12.5;

> TR:=p*qd;

> plot(TR, qd=0..40);

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> MR:=diff(TR,qd);

> solve(MR=0,qd);

> eval(p,qd=25);

We can show the price p=6.25 and the demand curve together in the following graph. Note that the price line is red at $6.25.

> plot({-.25*qd+12.5,6.25},qd=0..40);

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Find demand elastcity at the point of mamimum total revenue. This can be done by evalauting the following equation:

Demand elastcity: =[delta(q)/q]/[delta(p)/p] which simplifies to: [dq/dp]*[p/q]. The first term is the derivative of the demand function and the second term is the ratio of price over quantity. Substitution for the two terms yields:

-4*(6.25)/25)=1.

suppose the firm has a cost of production given by: TC:= 40 +3*qd. Find the profit maximizing output, and demand elastcity at that point. Profits:= Total Revenue - Total Cost.

> Restart;

Restart

> p:=12.5-0.25*qd;

p := -.25*qd+12.5

> TR:=(-.25*qd+12.5)*qd;

TR := (-.25*qd+12.5)*qd

> Pr:=TR-40 +3*qd;

Pr := (-.25*qd+12.5)*qd-40+3*qd

> Mr:=diff(Pr,qd);

Mr := -.50*qd+15.5

> solve(Mr, qd);

31.

> eval(p, qd=31);

4.75

> eval(TR, qd=31);

147.25

> plot({TR,Pr},qd=0..40);

Please note that the profit function --green curve reaches a maximum when qd=31, which is below the maximum point on the TR function. This is, of course, due to the cost constraint. Have fun!