Midterm, Part 2

The Problem:

Go to Ask Dr. Math on the Math Forum web site. Search the Archives for your first name. Browse some of the questions and answers that this search brings up. (If searching for you first name yields no results, try searching for the names of your relatives--this is just to get a random sampling of questions and answers from the archive--however, if you do use a different name, state what you used on your write-up).

Create a web page on which you can post the following:

a) Find a question for which the answer is particularly good. Provide a link to it from your page. Then explain what it is about the answer that makes it so good.

b) Find a question for which the answer is particularly poor. Provide a link to it from your page. Then explain what it is about the answer that makes it so bad. THEN, answer the question yourself (naturally you will want to answer it in a way you would consider 'good'), as if you were responding to the original e-mail.

Part A:

Reversing the Inequality

Dr. Peterson's explanation to Jonathan Lawrence Dulaney about why the direction of the inequality changes when multiplying by a negative number is especially good.

Dr. Peterson gave two explanations of the problem. The first exaplanation was very intuitive and very visual. It was extremely easy to understand. In fact, I wish that it had been explained to me this way when I was learning this material. His "inequality scale" method is easy to remember and makes very logical sense. Not only did Dr. Peterson express this explanation in words, he was very inventive and managed to create a very nice picture of the scale using the slash (/) character.

In addition to providing the scale example, Dr. Peterson provided a more mathematical approach for Jonathan. He explained the reversing of the inequality by using a number line and "reflect[ing] points through the origin" when the inequality is multiplied by a negative number. Once again, he provided a very ingenius illustration of the number line by using hyphens and plus signs (---+---+---+---).

Dr. Peterson provided Jonathan with two explanations of the problem. It wouldn't matter if Jonathan were a more visual learner (the scale approach) or a more mathematical learner (the number line approach), he would likely still be able to understand this problem. By providing two distinct ways of explaining this problem, and by using very inventive methods to illustrate the problem, Dr. Peterson did an outstanding job explaining this question to Jonathan.

Part B:

Minimizing the Cost of a Box

Dr. Jonathan's explanation of the Minimizing the Cost of a Box problem is insufficient. He explains one method of solving the problem, but only outlines it for the student. In my opinion, Dr. Jonathan skirted the most difficult part of the problem: setting up the equations. I feel that most Calculus students would realize that this problem would need to be solved using a derivative to find the minimum of a cost function, but probably the most difficult part of the problem is to actually find out what the cost function is. It is obvious that Dr. Jonathan did not actually solve the problem, since he states that "I have a feeling your problem will require the minimization of a function of two variables." This problem did not require minimizing a function of two variables.

Here is the solution I would have given to this student:

Since we want to find the minimum cost of this box, we will need to find a function that will give us the cost of the box in terms of the diminsions of the box. Once we have done that, we will need to find its minimum value (either by using technology or Calculus). So, let's set up some equations....

The volume of the box: lwh = 252
Since the box has a square base, l = w --> l2h = 252

The area of the top of the box: At = l2
The area of the bottom of the box: Ab = l2
The area of the four sides of the box: As = 4lh

The cost of the top of the box: Ct = 2At = 2l2
The cost of the bottom of the box: Cb = 5Ab = 5l2
The cost of the four sides of the box: Cs = 3As = 12lh
The total cost of the box: C = Ct + Cb + Cs = 2l2 + 5l2 + 12lh
Simplifying gives us: C = 7l2 + 12lh

Now let's solve our volume equation, l2h = 252, for h:
h = 252l-2

Now we can substitute for h in our total cost equation, C = 7l2 + 12lh:
C = 7l2 + 12l(252l-2)
Simplifiying gives us C = 7l2 + 3024l-1

So this is our final cost function:
C = 7l2 + 3024l-1

Graphing technology would be very good to use at this point. Let's graph this function, so we can get a look at it.

Graph: C = 7l2 + 3024l-1

It appears that the minimum of this function is between 5 and 10. So, let's find the actual minimum using Calculus. We know that to find a critical point of a function (in this case a minimum), we take the derivative of the function and find where it crosses the x-axis.

The derivative of C is
C' = 14l - 3024l-2

Now you should set C equal to 0 and solve the function for l. First, let's graph it, so we can get an idea of where it will cross the x-axis.

Graph: C' = 14l - 3024l-2

Solving this equation is a lot easier than it looks....

0 = 14l - 3024l-2
0 = 14l - 3024/l2
3024/l2 = 14l
3024 = 14l3
216 = l3
6 = l

So it appears that the length diminsion is 6 feet. That value does appear to coinside with both of our graphs. Now all we need to do is substitute our l value into the h = 252l-2. When we do that, we get h = 7. So the diminsions which minimize the cost of the box are a 6 foot by 6 foot base, with a 7 foot height.

An obvious extension of this problem would be to find out what the minimum cost actually is. It's not too difficult: can you do it?