# Application of quadratic functions to measurement and graphing

thanks for any help!

Q1. Find the equation of the surface area function of a cylindrical grain silo. The input variable is the radius (r). (the equation is to be graphed using a graphics calculator in the following question)

Height (h) = 5 meters

Surface Area (S)- unknown

Pi (p) = 3.142

So far I have:

S = 2pr^2 + 2prh (surface area formula)

S = 2p(r^2+5r)

S = 2pr(r+5)

S= 6.284r(r+5)

I am not sure if this is an equation I can use to answer Q2 Use the graphic calculator emulator to draw the equation obtained at Q1.

I have also come up with: 2pr^2 + 2prh + 0 (in the quadratic expression ax^2 + bx + c=0)

When I substitute values for r I get the same surface area for both equations but am not sure if I am on the right track!

Thank you for any help!

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(SA = Surface Area)

• SA (silo) = SA (cylinder) + $\frac{1}{2}$ SA (sphere)
• SA (cylinder) = $2\pi r h$
• SA (sphere) = $4\pi r^2$

So we have,

SA (silo) = SA (cylinder) + $\frac{1}{2}$ SA (sphere) = $2\pi r h + \frac{1}{2}4\pi r^2 = 2\pi r h + 2 \pi r^2 = 2 ~\pi~ r(h + r) = 2 ~\pi~ r(5 + r)$

Plot:

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Wow! thank you very much. This is the answer to Q3! –  David Apr 25 '13 at 5:27
In Q1 the silo is just a closed cylinder. (area of 2 circles + area of rectangle). In Q3 we invent the hemisphere on the bottom. So my first formula is on the right track? –  David Apr 25 '13 at 5:40
@David: Maybe it is me, but I am not seeing a Q3 in your original problem, so it is not clear to me what you are asking. Regards –  Amzoti Apr 25 '13 at 14:34
I'm not having too great of a day :-( –  amWhy Apr 26 '13 at 0:25
Me too, been several days like that for me! :-( Wish some fun problems would show up. –  Amzoti Apr 26 '13 at 0:26