# Math Fireworks

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• Topic: Binary relation, Volume, Cylinder
• Pages : 2 (434 words )
• Published : December 4, 2006

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October 26, 2006

Lakeside Explosives
22 Autumn Lane
Osgo, UV 41111

Dear Lakeside Explosives,

Here at Capstone it is our pleasure to help you with your new N-79 firecracker, hoping to give your customers truly more bang for their buck'.

To begin with I will tell you what the radius of the cylinder on the top of the N-79 was calculated to be and then will go through how we found this number. The radius we found was 1.4154683 cm, or 1.4 cm rounded to the nearest decimal point, the interval was found between 1.4 cm and 1.5 cm.

To begin with we had to determine what the total volume (cylinder + half a sphere) would be. Given the information that you at Lakeside will being using 10 grams of powder, and the powder you are using has a mass of .2g/cm³ we found the total volume to be 50 cm³.

10 grams * (1 cm³ / .2 grams) = 50 cm³

From there we had to form an equation to find the total volume of the half sphere and the cylinder in order to solve for a variable (x).

7חx² + ½ (4/3 ח x³) = 50 cm³

(Volume of cylinder + ½ volume of cylinder = total volume)

Using this equation we simplified the equation so that it would equal zero and we could make it a function that we could look at graphically. Thus we derived the equation:

7חx² + ½ (4/3 ח x³) - 50 cm³ = 0 = F(X)
We then used a graphing utility to plot the points and find out where the graph crosses the x axis (where y = 0, in coordinate (x, y)). It was shown that this occurs when x is between 1 and 2. From there we made a table with values, starting at 1 and increasing by .1 until we reached 2 as shown in the following figure:

xy
1.0-25.91
1.1-20.6
1.2-14.71
1.3-8.234
1.4-1.15
1.56.5487
1.614.876
1.723.844
1.833.466
1.943.754
2.054.72

We used the graphing utility to solve for the exact value, and found it to be 1.4154683 cm.

Again it has been a pleasure to work with you on your new firework endeavourer. If you have any...