> y:=x^3-3*x^2-24*x;

> y1:=diff(y,x);

> solve(y1,x);

Determine if the above function is convex or concave using the definition on page 65 of the text. Take second derivative and evaluate at x=-2, and x=4.

> y2:=diff(y1,x);

> eval(y2,x=-2);

The second derivative at x=-2 is negative, implying that the function is concave. Now evaluate at x=4.

> eval(y2, x=4);

The second derivative at x=4 is positive, implying that the function is convex. Now let us graph the function and verify the above the results.

> plot(y, x=-10..10);

> eval(y,x=-2);

> eval(y,x=4);

> plot({y,28,-80},x=-4..8);

>

If the second derivative is negative, a relative maximum has been found. Conversely, if the second derivative is positive, a relative minimum has been detected. When the second derivative is equal to zero, we have found an inflection point.

> solve(y2,x);

> eval(y1,x=1);

The Difference between optimization and maximimization? Optimization without constraints = maximization.

An Example: Optimize the following total revenue (Tr) function-- find critical values and test for max or min points.

> Tr:=2*q^3-30*q^2+126*q+59;

> Tr1:=diff(Tr,q);

> solve(Tr1, q);

> Tr2:=diff(Tr1,q);

> eval(Tr2, q=3);

> eval(Tr2, q=7);

> eval(Tr,q=3);

> eval(Tr,q=7);

> plot({Tr,221,157},q=0..10);

>