> c:=2*q^3-3*q^2+400*q+5000;

> tr:=4000*q-33*q^2;

> plot({c,tr},q=10..60);

> profits:=tr-c;

> plot(profits,q=0..40);

The first derivative of the profit function:

> dpr:=diff(profits,q);

Set the first derivative of the profit function equal to zero andf solve for q.

> solve(dpr,q);

There are two solutions: q=-30, and q=20. Only the second answer is acceptable because q>0.

Take the second derivative to see which point results in a maximum.

> ddpr:=diff(dpr,q);

> eval(ddpr,q=20);

Since the second derivative is negative at q=20, we have a maximum, and the function is CONCAVE .

The relationship among Total, Marginal and Average Product.

> TP:=90*k^2-k^3;

> MR:=diff(TP,k);

> AR:=TP/k;

> plot({MR,AR},k=0..70);

AS long as the MR curve is above the AR curve, the AR curve will be riing..More formally, take the derivative of both the MR and the AR functions, finds the maximum points. Note that the MR function reaches a maximum before AR does. See below.

> dMR:=diff(MR,k);

> solve(dMR, k);

> dAR:=diff(AR,k);

> solve(dAR,k);

Note that MR reaches a maximum at k=30, and AR reaches a maximum when k=45. Below is graph of the production function.

> plot(TP, k=0..70);

Note that when TP is at maximum MR=0. Solve the MR function with respect to k to detremine the exact value for the variable capital. The value at TP at the point of maximum= 60. Further note that the MR function is simply the first derivative of the TP function.

> solve(MR, k);

>