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This is the second part of a series of projects intended for first year calculus students. In this series, students will learn how to use different mathematical equations to assess the properties of solids and containers. My apologies if the "X" and "x" thing is confusing, but in some situations formatting makes using an asterisk impossible, so big X is a variable, and little x is multiply.

In this part of the project, students will learn how to use derivatives to find what value(s) of X gives us the maximum volume of a lidless box.

# Print Settings

**Printer: **

CraftBot

**Rafts: **

No

**Supports: **

No

**Resolution: **

optimum

**Infill: **

30% square

**Notes: **

I optimized this thing to be printed in ABS with maximum settings. I would recommend a skirt with an offset of 0mm (otherwise known as a brim) with 2-5 loops for ABS. I would also suggest an increased infill ratio.

# How I Designed This

For this project I used a free program from Autodesk called TinkerCAD. I have long been a user of AutoCAD which is a desktop drafting program from the same company. I used TinkerCAD to make it easier for educators and learners to be able to more easily replicate what I did. TinkerCAD is awesome because it's not only free, but it's easy to use. It also runs in your browser and has built in instructional lessons.

www.tinkercad.com

For part 6 of this project I started out with our product from part 2. If you remember, we used X=20mm which gave us a base of 60x20mm, two flaps with dimensions of 60x20mm, and two flaps with dimensions of 20x20mm. We then made this into a mold by adding a channel around the base and extending the flaps.

Part 2:

http://www.thingiverse.com/thing:1371700

# Project: Using derivatives

**Objective**

At the end of this lesson students will have a solid basic understanding of derivatives, what they represent, and how to find them.

**Audience**

This project is intended for new calculus students who have not yet learned about derivatives.

**Preparation**

If you and your students have completed steps 1-5, there is no more preparation that you should need to do. If you have not, then at least give the lessons a look. All of the math, equations, and conclusions are there. The other parts are listed as having this part as a derivative (get it?).

**Step 1: Review**

In order to get to the point where we can use derivatives, we need to first have a function to derive. Luckily, we already have an equation for volume as a function of X, from part 1 of this project. Another way to say this is f(x). Therefore, we have:

f(X) = (100 - 2X)(60 - 2X) * X

**Step 2: Simplify**

f(X) = (100 - 2X)(60 - 2X) * X
= (6000 - 200X -120X +4X^2) * X

= 4X^3 - 320X^2 + 6000x

**Step 3: Explain derivatives**

For simplicity sake and as it relates to the problem at hand, a derivative is equal to the slope of f(X) at any given point of X. I.e. if the you were to pick the point X = 5 on the graph of f(X), and the slope of the graph at that point were 2, then on the graph of DY/DX where X was equal to 5, Y would be equal to 2.

**Step 4: Find derivative**

There is a long version of finding derivatives and creating proofs that most teachers get a thrill of putting their students through. I am not one of those, so I will jump straight to the short version. To find a derivative, use this formula:

f(X) = n^r

DV/DX = rn^(r-1)

Therefore we have:

f(X) = 4X^3 - 320X^2 + 6000x

So....

DV/DX = 12X^2 - 640X + 6000

**Step 5: Find extreme values**

So let's think about what we need to find out, and what we have learned. We need to find out where V, or f(X), is greatest. We know that DV/DX is equal to the slope of f(X). If you think back to the graph we made in part 5, it appeared as though the slope at the greatest value of V was 0. Therefore, in order to find 'critical points' on a graph of f(X), we should be able to set DV/DX equal to 0. Since I was easy with the derivative, I made this part messy. Let's try it, the long way:

DV/DX = 12X^2 - 640X + 6000

12X^2 - 640X + 6000 = 0

6000 + -640X + 12X^2 = 0

Factor out the Greatest Common Factor (GCF), '4'.

4(1500 + -160X + 3X2) = 0

Ignore the factor 4.

1500 + -160X + 3X2 = 0

Begin completing the square. Divide all terms by

3 the coefficient of the squared term:

500 + -53.33333333X + X2 = 0

Move the constant term to the right:

500 + -53.33333333X + -500 + X2 = 0 + -500

Reorder the terms:

500 + -500 + -53.33333333X + X2 = 0 + -500

Combine like terms: 500 + -500 = 0

0 + -53.33333333X + X2 = 0 + -500

The X term is -53.33333333X. Take half its coefficient (-26.66666667).

Square it (711.1111113) and add it to both sides.

Add '711.1111113' to each side of the equation.

-53.33333333X + 711.1111113 + X2 = -500 + 711.1111113

Reorder the terms:

711.1111113 + -53.33333333X + X2 = -500 + 711.1111113

Combine like terms: -500 + 711.1111113 = 211.1111113

711.1111113 + -53.33333333X + X2 = 211.1111113

Factor a perfect square on the left side:

(X + -26.66666667)(X + -26.66666667) = 211.1111113

Calculate the square root of the right side: 14.529663152

Break this problem into two subproblems by setting

(X + -26.66666667) equal to 14.529663152 and -14.529663152.

Subproblem 1

X + -26.66666667 = 14.529663152

X = 14.529663152 + 26.66666667

X = 41.196329822

Subproblem 2

X + -26.66666667 = -14.529663152

X = -14.529663152 + 26.66666667

X = 12.137003518

X = {41.196329822, 12.137003518}

**Step 6: Set X equal to given values**

Normally we would plug both of these numbers in to see if either or both is what we were looking for. However, since we know that anything above 30 will result in a negative number, we will just use 12.137003518.

(100 - 2X)(60 - 2X) * X

(100-2(12.137003518))(60-2(12.137003518))12.137003518

32835.28

**Results**

Just as your students should have predicted in part 5, the greatest value of V was around X = 12. The greatest value of V was close to 32832 as well. Students will now be able to see just how handy, not to mention cool, derivatives can be.