To find absolute maxima and minima of a continuous function on a closed interval, we need to investigate the value of the function at the endpoints, as well as the value at any critical numbers of the function in the interior of the interval. A graph of the function can often give us a quick estimate of the maximum and minimum values and the x values at which they occur, while calculus provides us with the tools to find the exact values.  We study the function .

> f:=x->x-2*sin(x);

> plot(f,0..2*Pi);

By clicking on the points where the apparent minimum and maximum occur, estimate the coordinates of these two points.  Next, note that the minimum and maximum values appear to occur at interior points of the interval [ ], so they must correspond to critical numbers of the function. Thus to find these points, we solve for when the derivative is zero. (Note that the derivative of our function is defined at all values of x .)

> solve(D(f)(x)=0,x);

Note that Maple only found the x value corresponding to the local minimum. Let us compute the derivative, and set it equal to zero by hand.

> D(f)(x);

The derivative vanishes when , which occurs in our interval when , and also when . These are the two critical numbers. Once we know them, it is simple to have Maple evaluate the values of the function at these numbers and at the endpoints.

> f(0),f(Pi/3),f(5*Pi/3),f(2*Pi);

Of course, to decide which is largest, we need to estimate these values.

> evalf(f(0)),evalf(f(Pi/3)),evalf(f(5*Pi/3)),evalf(f(2*Pi));

It is clear that the second value is the absolute minimum, and the third is the absolute maximum of the function in the interval [ ].

Submission:

For the following problems

(a) Use a graph to estimate the absolute maximum and minimum values of the function to two decimal places.

(b) Use calculus to find the exact maximum and minimum values.

1. , where x is in [ ].

2. , where x is in [ ].

Submission worksheet:

A critical number is defined to be a value c in the domain of the function f at which either the derivative is zero or fails to exist. Solving for points at which the derivative is zero can be easy with Maple, if it is able to solve the equation . In the previous activity, we looked at an example in which Maple was not successful in giving us the complete solution to this problem. In this activity, we study how to find critical values at which the derivative fails to exist. Consider the function below.

> f:=x->x+abs(x^2-3*x-1);

Let us plot the function on the interval [ ] to see what it reveals about the relative and absolute extrema on the interval.

> plot(f,-2..4);

It is very apparent from the graph that there are two points at which the derivative does not appear to exist. Estimate the values from the graph. Then use the graph to estimate the absolute maximum and minimum values of the function, and the x values at which these values occur.  How can we solve for the values exactly? In this case, let us try to puzzle out where the cusps occur by thinking about what happens near these points. They correspond to the values of x at which the absolute value changes sign. Thus we can find them by solving for which values of x the expression inside the absolute value is equal to zero.

> solve(x^2-3*x-1);

> evalf(3/2+1/2*sqrt(13)),evalf(3/2-1/2*sqrt(13));

How close do these values correspond to the estimates from the graph? Note that once we have the x values for these critical points, we can find the corresponding y values easily. Thus finding the absolute maximum and minimum amounts to evaluating the four values below.

> f(-2),f(3/2+1/2*sqrt(13)),f(3/2-1/2*sqrt(13)),f(4);

Notice that the function has an absolute maximum of 7 at both endpoints. The absolute maximum or minimum value of the function can occur for more than one value of x . The absolute minimum value of the function is , and this minimum occurs when .

Submission:

For the function , find the critical values, the local extrema, and the absolute extrema in the interval [ ]. Note that you will need to find not only the values where the derivative does not exist, but also those where the derivative is zero. To solve for these values, use the methods of activity 1.

Submission worksheet:

Consider the function defined on the closed interval [-1,2] by

> f:=x->x^3-2*x^2+2*abs(x-1)+2;

In order to study the behavior of this function on the interval, let us first plot the function on the interval.

> plot(f,-1..2);

It appears that the function has a local maximum near , and a local minimum at . Also, it appears that the first point corresponds to a place where the derivative vanishes, and the second to a place where the derivative fails to exist.  Let us first solve for the point where the derivative is zero.

> solve(D(f)(x)=0,x);

> evalf(2/3-1/3*sqrt(10));

This value is close to the value that we estimated by hand.  It is easy to show that the derivative does not exist at x=1, by looking at the limit of the difference quotient. Let us define the difference quotient, and take its left and right limits at x=1, to see that they are different.

> dq:=(x,h)->(f(x+h)-f(x))/h;

> limit(dq(1,h),h=0,left);

> limit(dq(1,h),h=0,right);

> limit(dq(1,h),h=0);

The possible candidates for the extreme values of the function are the endpoints, the point where the derivative vanishes, and the point where the derivative fails to exist. Since this is only a set of 4 points, it is easy to evaluate which of them gives the absolute maximum and minimum values of the function on this domain.

> f(-1);f(2/3-1/3*sqrt(10));f(1);f(2);

> evalf(f(2/3-1/3*sqrt(10)));

We find the absolute minimum, which is also a relative minimum of the function, is 1, and this occurs at . The absolute maximum of the function is approximately 4.4, and this occurs when . Although in this case the extreme values did not occur at the endpoints, they often do, so don't forget to check them out.

Submission:

For the following functions:

(a) Plot the function, f, over the given interval to see its general behavior there. Also plot .

(b) Find the interior points where .

(c) Find the interior points where does not exist.

(d) Evaluate the function at all the points found in parts (b) and (c) and at the endpoints of the interval.

(e) Find the function's absolute extreme values on the interval and identify where they occur.

1. , [ ] .

2. , [ ]

3. , [ ]

4. , [ ]

Submission worksheet:

In applications of Calculus to problems in the sciences, it is not uncommon for the function to have exactly one critical value in the feasible domain of definition. In this case, the First Derivative Test for Absolute Extrema can usually be applied to find an absolute extreme for the function.  For example we can show when by finding the critical value for the function , which occurs at , and then showing that this critical value corresponded to a local minimum for the function on the interval ( ). The value , of the function is thus the minimum value on this interval, so the inequality above holds.

Submission:

(a) Demonstrate the necessary computations to show the above inequality is true.

(b) Find the point on the line that is closest to the origin (the point (0,0)). Hint: The distance formula is . Since and we have .  Be sure to show that the hypotheses of the First Derivative Test for Absolute Extrema are met before applying this criteria.

Submission worksheet:

Given the power of a graphing tool such as Maple, it might occur to you to wonder why we need such procedures as the First Derivative Test for Local Extrema , when it is easy to simply plot a curve and look to see where it reaches a maximum or minimum value. Of course, one problem with the method of plotting is that it does not tell us what is happening outside the window we look at. Sometimes we do not have the formula for a function, but do have one for its derivative, and in this case, the First Derivative Test for Local Extrema can tell us where the function reaches a local maximum, even when we don't have a formula.  For example, suppose that we are given the function f below, which is the derivative of an unknown function F .

> f:=x->(x-1)/(x^2+1);

How can we determine where the function F has a minimum value, when we do not know a formula for the function.  Clearly, there is only one zero for the derivative, and we can study the behavior of the derivative near this zero to apply the First Derivative Test for Local Extrema . Let us plot .

> plot(f(x),x);

Applying the First Derivative Test, we see that since the derivative changes sign from negative to positive at , our unknown function must have a local minimum value there. Notice that the derivative is negative for all , and is positive for all , so in fact, the function is decreasing up until , and increasing thereafter, so the value of the function at is the absolute minimum of the function. We cannot tell what value this is, because the derivative never reveals the value of a function at a point.  Maple can sometimes solve for a function whose derivative is a specified function. This process is called anti-differentiation, and will be studied extensively in math 215. Just for concreteness here, we show how to use the int command to find a function whose derivative is our function , and plot its graph to see that it really has a minimum at .

> F(x):=int(f(x),x);

> plot(F(x),x);

Submission:

Suppose that you know that the derivative of the function F you are interested in is given by  .  Use the First Derivative Test for Local Extrema to classify the local extrema of the function. Then use the method above to find a function F whose derivative is given by the function above, and graph it along with its derivative on the same plot. Explain the relationship between the graphs of the function and its derivative.

Submission worksheet:

In a beehive, each cell is a regular hexagonal prism, open at one end with a trihedral angle at the other end. It is believed that bees form their cells in such a way as to minimize the surface area for a given volume, thus using the least amount of wax in cell construction, Examination of these cells has shown that the measure of the apex angle theta is amazingly consistent. Based on the geometry of the cell it can be shown that the surface area is given by  where s , the length of the sides of the hexagon, and h , the height, are constant.

Submission:

(a) Calculate .

(b) What angle should be bees prefer?

(c) Determine the minimum surface area of the cell (in terms of s and h ).

(d) pick , and and plot the function . Choose other values of s and h and plot the function.

Submission worksheet:

Consider following directions to make a box with a lid from a sheet of cardboard that is 24 in by 36 in. First fold the cardboard in half to form a 24 by 18 rectangle, and then from each corner cut out an x by x square. When you then unfold the cardboard sheet, you can assemble a box. It's a bit of a chore, but we can use Maple to draw a picture of this situation.

> with(plots):with(plottools):

Warning, the name changecoords has been redefined

Warning, the name arrow has been redefined

> p1:=polygon([[0,0],[6,0],[6,4],[0,4]],color=tan):

> p11:=polygon([[-0.8,0],[-0.4,0],[-0.6,0],[-0.6,1.6],[-0.6,0]]):

> p12:=polygon([[-0.8,4],[-0.4,4],[-0.6,4],[-0.6,2.4],[-0.6,4]]):

> p13:=polygon([[0,-0.3],[0,-0.7],[0,-0.5],[2,-0.5],[0,-0.5]]):

> p14:=polygon([[6,-0.3],[6,-0.7],[6,-0.5],[3.8,-0.5],[6,-0.5]]):

> l1:=textplot([[-0.7,2,`24 in.`],[3,-0.5,`36 in.`]]):

> p2:=polygon([[8.5,0.5],[8.5,0],[10.5,0],[10.5,0.5],[11,0.5],[11,3.5],[10.5,3.5],[10.5,4],[8.5,4],[8.5,3.5],[8,3.5],[8,0.5]],color=tan):

> p3:=polygon([[14.5,0.5],[14.5,0],[16.5,0],[16.5,0.5],[17,0.5],[17,3.5],[16.5,3.5],[16.5,4],[14.5,4],[14.5,3.5],[14,3.5],[14,0.5]],color=tan):

> p4:=polygon([[17.5,0.5],[17.5,0],[19.5,0],[19.5,0.5],[20,0.5],[20,3.5],[19.5,3.5],[19.5,4],[17.5,4],[17.5,3.5],[17,3.5],[17,0.5]],color=tan):

> l2:=textplot([[9.65,-0.5,`18 in.`],[17,-0.5,`36 in.`]]):

> p15:=polygon([[8,-0.3],[8,-0.7],[8,-0.5],[8.5,-0.5],[8,-0.5]]):

> p16:=polygon([[14,-0.3],[14,-0.7],[14,-0.5],[16,-0.5],[14,-0.5]]):

> p17:=polygon([[11,-0.3],[11,-0.7],[11,-0.5],[10.5,-0.5],[11,-0.5]]):

> p18:=polygon([[20,-0.3],[20,-0.7],[20,-0.5],[18,-0.5],[20,-0.5]]):

> l3:=textplot([[8.35,1,`x`],[8.35,3.2,`x`],[8.75,.4,`x`],[8.75,3.8,`x`],[10.2,.4,`x`],[10.2,3.8,`x`],[10.7,1,`x`],[10.7,3.2,`x`]]):

> display(p4,p3,p2,p1,p11,p12,p13,p14,p15,p16,p17,p18,l1,l2,l3,scaling=constrained,axes=none);

Submission:

(a) Write a formula V(x) for the volume of the box.

(b) Find the domain of V for the problem situation and graph V over this domain.

(c) Use a graphical method to find the maximum volume and the value of x that gives it.

(d) Confirm your result in (c) analytically.

(e) Find a value of x that yields a volume of 1120 cubic inches.

(f) Write a paragraph describing the issues that arise in part (b).

Submission Worksheet: