The surface area of a soup can represents the total area of the can's exterior and may be found by the formula S.A. = 2πrh + 2πr2. A soup company designing its own cans wants to determine the height of its cans with a given surface area and radius. Help the company by solving the surface area formula for h.
So which part of the right hand side of the equation contains the h?
2pirh
yes :) So from the last problem we did, we learned to ut all the stuff that doesn't contain what you're trying to solve for onto the other side of the equation, right? So can you show me how you would put everything that doesnt have "h" in it to the other side with the appropriate sign?
and btw, pretend the "SA" is actually like a "0" so when you move the stuff onto the other side, don't include it in your problem :)
subtracting 2pir2 on both sides ?
You got it! :)
\[-2\pi r^2= 2\pi rh\]
Now what else can we do so we get "h" by itself?
divide ?
mmhmm, but divide what by what?
both sides by 2pirh ?
Wait,no no, remember,you are solving for h, if you divide with it, you cannot solve for it!
oh so -2pir^2
We currently have \[-2\pi r^2=2\pi rh\] in order to get "h" by itself, we need to divide by something, but we're not going to touch the h, because that is what we are solving for. what other things can we divide by to get H all by itself?
2pir ?
yessir :)
So we will have \[-\frac{ 2\pi r^2 }{ 2\pi r }=h\]correct?
yes
do we have to do something else ?
Yes :) Now you can reduce the function of h. Notice how there are like terms (similar terms such as 2, pi and r) in both the numerator and denominator? We can eliminate those as well
what do we get ?
We have \[(-\frac{ 2 }{ 2 })(\frac{ \pi }{ \pi })(\frac{ r^2 }{ r })=h\] so we can reduce them all.
i divide each ?
Yep!
pi divided by pi would be ?????? 1 ?
You got it :D
and how about the 2's?
-2 / 2 = -1
mmhmm. and the r's?
r^2
\[\frac{ r^2 }{ r }= r ^{2-1}\] can you reduce that?
r
And now we put all the values together :)
-11r ?
\[(-1)(1)(r)=h\]
thats the answer ? cuz i have different answer choices
Lets see. I'll rework the problem out here for you to check your work?
\[h = \frac{ 2\pi r }{ S.A. + 2\pi r }\] \[h = \frac{ 2\pi r }{ S.A. - 2\pi r }\] \[h = \frac{ S.A. + 2\pi r }{ 2\pi r }\] \[h = \frac{ S.A. - 2\pi r }{ 2 \pi r }\]
Or wait... I think we had to include SA as part of the answer. but no worries, it's all going to be the same except now we're going to have SA as part of the answer.
yes! :)
all the other steps we had done were correct, we just have to reinclude SA
So i will work out the problem too, and you can check your work with mine?
ok :)
\[SA=2\pi rh+2\pi r^2\]\[SA-2\pi r^2=2\pi rh\]\[\frac{ SA-2\pi r^2 }{ 2\pi r }=h\]
Now i believe we cannot eliminate any variables because there are no common terms that are shared with SA. So that is our answer. Before when we did not include SA as part of our answer, we were able to reduce all the terms because they shared stuff in common
so which one on top would it be ?
I am sorry, i made a mistake in not including SA into solving this problem.
its ok... i sent u my choices above ? i still dont know which one since none of my choices have ^2
Hmm... it's not matching any of these answers. let's see \[h=\frac{ SA }{ 2\pi r }-\frac{ 2\pi r^2 }{ 2\pi r}\] this would reduce down to \[h=\frac{ SA }{ 2\pi r }-r\]
the closest thing is choice D, but that's missing a square over the r the other choices don't even come close to the true answer of \[h = \frac{ S.A. - 2\pi r^2 }{ 2 \pi r }\] which is what Jhannybean got above
so my guess is that there's a typo
is there a typo ? here is my question
\[h=\frac{S.A-2\pi r^2}{2\pi r}\]
ill just put D :p
There's no answer, lol
hmmm
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