Calculus Video: Optimization (Max-Min) - NEW Improved Version!

Well, I felt badly about that error I made in the original version from a few days ago (which will now be deleted). I also decided to change some portions, electing to solve for the critical values using algebra rather than by the graphing calculator. Finally, I took a tighter view of the whiteboard so that the writing will appear larger. There will be some glare on the board which I hope will not be too distracting. I hope you will find this more helpful and again I apologize for any confusion caused. If you stored the original video, I would ask you to delete that.

HAPPY HOLIDAYS!

The problem in the video below demonstrates important concepts as well as the standard procedure for solving optimization problems. There is also a brief discussion of a heuristic I have found very useful when teaching these kinds of applications. As always I depend on you to share your thoughts. I keep saying this knowing there might not be too many comments!

Solutions and Discussion re Squares problem

Since I only received one comment from the squares problem, I guess readers were either bored by this question or are awaiting my solution(s)!

Solution I: Divide side AB into segments of lengths 2+x, 2-x. Here, x can be any real between -2 and 2. Similarly, divide BC into segments of lengths 2-y and 2+y. One can demonstrate that the order here is irrelevant. Then the product of the areas of either pair of non-adjacent rectangles can be expressed (after rearrangement of factors) as
(2+x)(2-x)(2+y)(2-y) = (4-x²)(4-y²).
From the restrictions on x and y, it follows that x² is greater than or equal to zero and less than 4. Similarly for y.
Therefore, (4-x²)(4-y²) ≤ 4⋅4 or 16.
This also demonstrates that the maximum product occurs when x=0 and y=0! QED

Solution II (using the Arithmetic-Geometric Mean Inequality): We will prove a general result for squares of side m. This will be forthcoming and there may be a visual surprise! Stay tuned!

Just Another Square Problem? A Means to an End...

Before announcing the thousands (or less!) winners of the Name That Mathematician Challenge, I came across a problem about dissecting a square ABCD with lines PQ and RS which are parallel to the sides of the square. (see diagram).

Naturally, I decided to make it into a deeper investigation. Students and/or readers will be asked to find the maximum value of the product of the areas of either pair of non-adjacent rectangles formed. There are many approaches here, one of which uses the famous Arithmetic Mean-Geometric Inequality. As usual you will work from the particular to the general, beginning with a specific value for the sides of ABCD.

STUDENT/READER INVESTIGATION - PART I
The given conditions about the diagram are given above.

For Part I, we will assume each side of the square has length 4.

(1) (Particular) If AP = 3 and RC = 2, determine the product of the areas of APTS and RTQC. Do the same for the other pair of non-adjacent rectangles formed. Do you believe this product is the maximum possible as we vary the positions of segments PQ and RS?

(2) (General) Show that the product of the areas of either pair of non-adjacent rectangles formed is less than or equal to 16. For example the product of the areas of APTS and RTQC is ≤ 16.

Notes:
(1) Do you think many students would guess what the configuration would be for the maximum product to occur? Is proving the conjecture much more difficult?
(2) The challenge here is to find an effective use of variables to denote the segments. There are many possibilities, some much more efficient than others.
(3) I will add additional parts to this challenge after receiving comments on Part I. How would you generalize this result further? More interestingly, there is a way to prove Part I using the AM-GM Inequality?

The Classic Cone Inscribed in the Sphere Problem: Developing Relationships Before Calculus

Update: View the series of videos here explaining the procedure for solving the cone in the sphere problem below as well as related questions.

Many Algebra 2 and Precalculus textbooks have begun to include those challenging 3-dimensional geometry questions involving 2 or more variables and/or constants. However, we know from the difficulty that calculus students continue to have with these, that we need to do more before students do their first optimization problems in calculus. You know the kind: Determine the radius of the __________ of maximum volume that can be inscribed in a _________ of radius R. These problems have fallen out of favor somewhat with the AP Development Committee, perhaps because they lack that real-world flavor or perhaps because they had become predictable or perhaps too hard. I would argue they have been part of the rites of passage for calc students for many generations for a reason - they blended the spatial reasoning of geometry with the need to identify variable relationships and reduce the number of conditions down to one function of one variable if possible. In other words, they help to develop mathematical sophistication. I 'cut my teeth' on these -- did you? Any calculus teachers reaching this topic yet in AP Calc?

Anyway here's an activity for you Algebra 2 or Precalculus students to prepare them for these challenges. As usual we proceed from the concrete (i.e., given numerical dimensions) to the abstract. Rather than attempt to draw the diagram, which is fairly challenging for me given the tools I have, I will describe the problem verbally. Good luck!

STUDENT ACTIVITY

(1) A right circular cone of height 32 is inscribed in a sphere of diameter 40.
Note: Students need to learn how to make a diagram of this problem situation.

(a) Determine the radius of the cone.
(b) Determine the volume of the cone. [Imagine asking students to memorize the formula!]
(c) Keep the diameter of the sphere at 40. This time, determine both the radius and volume of the inscribed cone whose height is 80/3. The numbers are messy but try to work in exact form (fractions, radicals) before rushing to the calculator to convert everything to decimals. Oh well, we all know what will happen here!
(d) Try another value for the height of the cone, keeping the diameter of the sphere at 40. See if you can produce a volume greater than in (c). Any conjectures?

(2) We could throw in an intermediate step by using a parameter R to denote the radius of the sphere, and use numerical values for different possible heights of the cone, but I'll leave that to the instructor. Instead, we'll jump to the abstract generalization:

A right circular cone of height h is inscribed in a sphere of radius R.
(a) Express the radius, r, of the cone in terms of R and h.
(b) Express the volume, V, of the cone as function of h alone (R is a constant here).
(c) Use your expression for r and your function for V to verify your results in (1).
(d) Calculus Students: You know what the question will be! Oh, alright:
Determine the dimensions and volume of the right circular cone of maximum volume that can be inscribed in a sphere of radius R. Anything strike you as interesting in this result?