Diameter of the base of a cylinder

[geekily]

A right circular cylinder has a surface are of 112pi. f the height of the cylinder is 10, find the diameter of the base.

I'm in "mathematics for elementary teachers", so we're staying away from the more complicated formulas. All we learned for surface area of a cylinder is that it's the area of the rectangle (the middle part) plus the area of 2 circles. I set the area of the rectangle equal to 10 x 2pi(r) and each circle equal to pi(r)^2.

I set it up as 10 x 2pi(r) + (pi(r)^2 + pi(r)^2 = 112pi. I'm guessing there's something (well, a lot of things) wrong with my combining the terms, but I can't tell what it is:

20pi(r) + 2(pi(r)^2) = 112pi
r + 2(pi(r)^2) = 112pi/20pi
r + 2(pi(r)^2) = 5.6
r + 2pi + 2r^4 = 5.6
r + 2r^4 = 5.6 - 2pi
r^9 = 5.6 - 2pi
r = calculator error because it's a non-real number.

Please help!

Thank you!

 

[o_O]

It seems that you have a few calculation errors.

From your first to second line, you went from:
\(\displaystyle 20\pi r + 2\left[\pi r^{2}\right] = 112\pi\)

to

\(\displaystyle r + 2\left[\pi r^{2}\right] = \frac{112\pi}{20\pi}\)

When you divide a term from both sides, you must make sure that each term of each side is divided by it. i.e.) On the second line on the left hand side, you divided \(\displaystyle 20\pi r\) by \(\displaystyle 20 \pi r\) but what about \(\displaystyle 2 \pi r^{2}\)?

Also, from the 3rd to 4th line, you multiplied out \(\displaystyle 2\left(\pi r^{2}\right) to \(\displaystyle 2\pi + 2r^{4}\). How does this equate? If you multiplied out something like \(\displaystyle 2\left(\pi + r^{2}\right)\) you might get something close: \(\displaystyle = 2\pi + 2r^{2}\) but this is not the case. When dealing with the \(\displaystyle 2\left(\pi r^{2}\right)\), this is no different from \(\displaystyle 2\pi r^{2}\).

One thing that you seem to have a problem with (such as your 5th to 6th line) is that when dealing with coefficients, they do not apply to exponents. For example:
\(\displaystyle 2r^{4} \neq r^{8}\). Also, you cannot ADD terms with the same bases but different exponents: ex.) \(\displaystyle r^{4} + r^{3} \neq r^{7}\).

I think you should review over your algebra when dealing with these types of things because when solving this particular question, you'll ultimately end up having to factor a polynomial which I don't know whether or not you can do it. Anyway, I hope this advice was helpful and perhaps someone else can do a better job of explaining.\)

 

[Mrspi]

A right circular cylinder has a surface are of 112pi. f the height of the cylinder is 10, find the diameter of the base.

I'm in "mathematics for elementary teachers", so we're staying away from the more complicated formulas. All we learned for surface area of a cylinder is that it's the area of the rectangle (the middle part) plus the area of 2 circles. I set the area of the rectangle equal to 10 x 2pi(r) and each circle equal to pi(r)^2.

I set it up as 10 x 2pi(r) + (pi(r)^2 + pi(r)^2 = 112pi. I'm guessing there's something (well, a lot of things) wrong with my combining the terms, but I can't tell what it is:

20pi(r) + 2(pi(r)^2) = 112pi
r + 2(pi(r)^2) = 112pi/20pi Whoops!!! You did NOT divide both sides by 20 pi!!
r + 2(pi(r)^2) = 5.6
r + 2pi + 2r^4 = 5.6
r + 2r^4 = 5.6 - 2pi
r^9 = 5.6 - 2pi
r = calculator error because it's a non-real number.

Please help!

Thank you!

20 r pi + 2 r<SUP>2</SUP>pi = 112 pi

Divide both sides by pi:

(20 r pi) / pi + (2 r<SUP>2</SUP> pi) / pi = (112 pi) / pi

20r + 2r<SUP>2</SUP> = 112

2r<SUP>2</SUP> + 20r - 112 = 0

You can divide both sides by 2.....

r<SUP>2</SUP> + 10r - 56 = 0

Now, solve using your favorite method for solving a quadratic (this DOES factor nicely!).

You'll get two solutions, but only one will be appropriate for the radius of a cylinder.

 

[galactus]

You made an algebra booboo. You divided when you shouldn't have.

\(\displaystyle \L\\2{\pi}r(10)+2{\pi}r^{2}=112{\pi}\)

\(\displaystyle \L\\20{\pi}r+2{\pi}r^{2}=112{\pi}\)

Divide by 2Pi:

\(\displaystyle \L\\10r+r^{2}=56\)

So, you have the quadratic to solve:

\(\displaystyle \L\\r^{2}+10r-56=0\)

Only one answer will be good. The positive one.

 

[geekily]

Ah, I thought my mistake might be something like that. I've never quite grasped what can and can't be canceled out and all of that. Thank you so much!

I got

(r + 14) (r - 4)

r = -14 and 4, so d = 8. I think that's right.

Thanks again, both of you!