1. ## solid mensuration

hello everyone !!
..I'm new here and i wanna ask some help...if its ok with you..

I'm having an assignment today and it really gives me a hard time answering this problem,...

"A cylindrical tin can holding 2 gal. has its height equal to the diameter of its base. Another cylindrical tin can with the same capacity has its height equal to twice the diameter of its base. Find the ratio required for making the two can with covers."

..well my book got the answer but doesn't have any solutions...

.
..btw this question comes from the "SOLID MENSURATION BY KERN AND BLAND" and i tried to look for answers in the internet...all i got is the "SOLID MENSURATION BY ASIN" but the sad thing is, he's writings were too small and i cant even see it clearly...and because this copy that i have is just a scan from another scan book...

hope you guys can help me

2. ## Re: solid mensuration

Hello, REDnight!

Welcome aboard!

A cylindrical tin can holding 2 gal. has its height equal to the diameter of its base.
Another cylindrical tin can with the same capacity has its height equal to twice the diameter of its base.
Find the ratio required for making the two can with covers.

The shorter can has radius $R$ and height $H = 2R.$
Its volume is: . $V_1 \:=\:\pi R^2H \:=\:\pi R^2(2R) \:=\:2\pi R^3$

The taller can has radius $r$ and height $h = 4r.$
Its volume is: . $V_2 \:=\:\pi r^2h \:=\:\pi r^2(4r) \:=\:4\pi r^3$

The volumes are equal: . $2\pi R^3 \:=\:4\pi r^3 \quad\Rightarrow\quad R^3 \:=\:2r^3 \quad\Rightarrow\quad \frac{R}{r} \:=\:\sqrt[3]{2}$ .[1]

The short can has surface area: . $A_1 \:=\:2\pi R^2 + 2\pi RH \:=\:2\pi R^2 + 2\pi R(2R) \:=\:6\pi R^2$

The tall can has surface area: . $A_2 \:=\:2\pi r^2 + 2\pi rh \:=\:2\pi r^2 + 2\pi r(4r) \:=\:10\pi r^2$

Ratio: . $\frac{A_1}{A_2} \:=\:\frac{6\pi R^2}{10\pi r^2} \:=\:\frac{3R^2}{5r^2} \:=\:\frac{3}{5}\left(\frac{R}{r}\right)^2$

Substitute [1]: . $\text{Ratio} \:=\:\frac{3}{5}\left(\sqrt[3]{2}\right)^2 \:=\:\frac{3\sqrt[3]{4}}{5}$