Tag Archives: optimization problems

Good Question 7 – 2009 AB 3

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Another in my occasional series on Good Questions to teach from. This is the Mighty Cable Company question from the 2009 AB Calculus exam, number 3

This question presented students with a different situation than had been seen before. It is a pretty standard “in-out” question, except that what was going in and out was money. Students were told that the Mighty Cable Company sold its cable for $120 per meter. They were also told that the cost of the cable varied with its distance from the starting end of the cable. Specifically, the cost of producing a portion of the cable x meters from the end is 6\sqrt{x} dollars per meter. Profit was defined as the difference between the money the company received for selling the cable minus the cost of producing the cable.  Steel Wire Rope 3

Students had a great deal of trouble answering this question. (The mean was 1.92 out of a possible 9 points. Fully, 36.9% of students earned no point; only 0.02% earned all 9 points.) This was probably because they had difficulty in interpreting the question and translating it into the proper mathematical terms and symbols. Since economic problems are not often seen on AP Calculus exams, students needed to be able to use the clues in the stem:

  • The $120 per meter is a rate. This should be deduced from the units: dollars per meter.
  • The cost of producing the portion of cable x meters from one end cable is also a rate for the same reason. In economics this is called the marginal cost; the students did not need to know this term.
  • The profit is an amount that is a function of x, the length of the cable.

Part (a): Students were required to find the profit from the sale of a 25-meter cable. This is an amount. As always, when asked for an amount, integrate a rate. In this case integrate the difference between the rate at which the cable sells and the cost of producing it.

\displaystyle P(25)=\int_{0}^{25}{\left( 120-6\sqrt{x} \right)dx}=\$2500

or

\displaystyle P(25)=120(25)-\int_{0}^{25}{6\sqrt{x}\ dx}=\$2500

Part (b): Students were asked to explain the meaning of \displaystyle \int_{25}^{30}{6\sqrt{x}dx} in the context of the problem. Since the answer is probably not immediately obvious, here is the reasoning involved.

This is the integral of a rate and therefore, gives the amount (of money) needed to manufacture the cable. This can be found by a unit analysis of the integrand: \displaystyle \frac{\text{dollars}}{\text{meter}}\cdot \text{meters}=\text{dollars} .

Let C be the cost of production, so \displaystyle \frac{dC}{dx}=6\sqrt{x}, and therefore, \displaystyle \int_{25}^{30}{6\sqrt{x}dx}=C\left( 30 \right)-C\left( 25 \right) by the Fundamental Theorem of Calculus (FTC).

Therefore, \displaystyle \int_{25}^{30}{6\sqrt{x}dx} is the difference in dollars between the cost of producing a cable 30 meters long, C(30), and the cost of producing a cable 25 meters long, C(25). (Another acceptable is that the integral is the cost in dollars of producing the last 5 meters of a 30 meter cable.)

Part (c): Students were asked to write an expression involving an integral that represents the profit on the sale of a cable k meters long.

Part (a) serves as a hint for this part of the question. Here the students should write the same expression as they wrote in (a) with the 25 replaced by k.

\displaystyle P(k)=\int_{0}^{k}{\left( 120-6\sqrt{x} \right)dx}

or

\displaystyle P(k)=120k-\int_{0}^{k}{6\sqrt{x}\ dx}

Part d: Students were required to find the maximum profit that can be earned by the sale of one cable and to justify their answer. Here they need to find when the rate of change of the profit (the marginal profit) changes from positive to negative.

Using the FTC to differentiate either of the answers in part (c) or by starting fresh from the given information:

\displaystyle \frac{dP}{dx}=120-6\sqrt{x}

\displaystyle \frac{dP}{dx}=0 when x = 400 and P(400)= $16,000.

Justification: The maximum profit on the sale of one cable is $16,000 for a cable 400 meters long. For 0<x<400,{P} '(x)>0 and for x>400,{P} '(x)<0 therefore, the maximum profit occurs at x = 400. (The First Derivative Test).

Once students are familiar with in-out questions, this is a good question to challenge them with. The actual calculus is not that difficult or unusual but concentrating on the translation of the unfamiliar context into symbols and calculus ideas is different. Show them how to read the hints in the problem such as the units.

Steel cable or steel wire rope as it is called also has some interesting geometry in its construction. You can find many good illustrations of this, such as the ones below, by Googling “steel wire rope.”

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Soda Cans

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A typical calculus optimization question asks you to find the dimensions of a cylindrical soda can with a fixed volume that has a minimum surface area (and therefore is cheaper to manufacture).

Let r be the radius of the cylinder and h be its height. The volume, V, is constant and V=\pi {{r}^{2}}h. The surface area including the top and bottom is given by

S=2\pi rh+2\pi {{r}^{2}}

Since \displaystyle h=\frac{V}{\pi {{r}^{2}}}, the surface area, S, can be expressed as

S=2V{{r}^{-1}}+2\pi {{r}^{2}}

To find the value of r that will give the smallest surface area we find the derivative, set it equal to zero and solve for r:

\displaystyle \frac{dS}{dr}=-2V{{r}^{-2}}+4\pi r

This will equal zero when \displaystyle r=\sqrt[3]{\frac{V}{2\pi }} and substituting into the expression above \displaystyle h=\sqrt[3]{\frac{4V}{\pi }}.

Then \displaystyle \frac{h}{r}=\sqrt[3]{\frac{\frac{4V}{\pi }}{\frac{V}{2\pi }}}=2, so h=2r. In the optimum can the height is equal to the diameter.

The thing is that very few cans, especially beverage cans are anywhere near this “square “ shape. The closest I could find in my pantry was a tomato sauce can holding 8 oz. or 277 mL. The inside dimensions are about 65 cm. by 75cm.  Compare this to the 12 oz. soda can holding 355 mL. The usual reason given for this departure from the mathematically best shape is the taller can is easier to hold especially for children.

IMG_0442

What got me interested in this was the video below. While there is no overt calculus mentioned, there is a lot of math. There are also STEM considerations, specifically engineering. As you watch look for the math and engineering ideas that are mentioned and discuss them with your class. Here are a few:

  1. Geometry: Why a cylinder? Why not a sphere or a cube?
  2. Engineering: When cutting circles out of rectangular sheets of aluminum there is a lot of unused metal. Why is all this waste not a problem? This goes to materials engineering; steel is more difficult to recycle than aluminum.
  3. Math: Efficient packing is also a consideration. Check the calculations in the video as to the most efficient way (least empty space) to pack containers. Why do they not use the most efficient?
  4. Geometry: The (spherical) dome is a very strong shape. In what other places are domes used? Why?
  5. Engineering: How does pressurizing the cans make them stronger?
  6. Geometry and Engineering: The elongated ridges on the sides of non-pressurized steel cans strengthen the sides. How are these ridges similar to the dome or circular arch?
  7. Physics: Look for a discussion of first- and second-class leavers.
  8. Engineering: What other advantages are there to using the very thin aluminum can.

At the end of the video 6 other videos are mentioned. These are also interesting and show the same process in cartoon form and in video of the machines making cans. The links to these are here:

Rexam: http://www.youtube.com/watch?v=7dK1VV…
How It’s Made: http://www.youtube.com/watch?v=V7Y0zA…
Anim1: https://www.youtube.com/watch?v=WU_iS…
Anim2:https://www.youtube.com/watch?v=hcsDx…
Drawing: https://www.youtube.com/watch?v=DF4v-…
Redrawing: http://www.youtube.com/watch?v=iUAijp…

Inequalities

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This is an “extra” post on a technique that students are using a lot right now in graphing functions and working on optimization problems.

In analyzing a derivative to find critical points and then the intervals where the function increases and decreases you need to solve these inequalities. I’ve observed many students solving inequalities the hard way.

By that I mean they will pick a number in each interval between the critical points and then substitute it into the derivative and do a lot of arithmetic to determine whether the derivative is positive or negative. Arithmetic is not necessary, when all you really need is the sign of the derivative. The method I suggest is easier and involves no arithmetic and therefore precludes any arithmetic mistakes

A complete discussion of this idea with examples click here or look on my Resources page.