# Minimization problem

• Dec 8th 2011, 03:10 AM
MathMinors
Minimization problem (multiple choice(
A cylindrical can is to contain 2000 cu. inches of liquid. What dimensions will minimize the amount of metal used in the construction of the can?
Choices:
a. D = 6.828in, H = 13.656
b. D = 6.424 in, H = 13.555
c. D = 6.222 in , H = 13.757
d. D = 6.626 in, H = 13.454

I honestly dont know where to start
• Dec 8th 2011, 03:44 AM
chisigma
Re: Minimization problem (multiple choice(
Setting r the radious of the base of the can and h its height, then we have...

$v= \pi\ r^{2}\ h$ (1)

$a= \pi\ r^{2} + 2 \pi r h$ (2)

From (1) You derive...

$h= \frac{v}{\pi r^{2}}$ (3)

... so that from (2) and (3) You obtain...

$a= \pi r^{2} + \frac{2 v}{r}$ (4)

Because v is known, Your task is to find the value of r that minimizes a...

Kind regards

$\chi$ $\sigma$
• Dec 8th 2011, 04:18 AM
MathMinors
Re: Minimization problem (multiple choice(
Quote:

Originally Posted by chisigma

$a= \pi\ r^{2} + 2 \pi r h$ (2)

circular area of bot and top included ?
$a= 2\pi\ r^{2} + 2 \pi r h$

thnks btw
• Dec 8th 2011, 07:32 AM
chisigma
Re: Minimization problem (multiple choice(
The cilinder has been considered 'open on the top'... if You want to include the top circular area the procedure is similar and the task is left to You...

Kind regards

$\chi$ $\sigma$
• Dec 9th 2011, 02:07 AM
chisigma
Re: Minimization problem (multiple choice(
Taking into account the top and bottom circular areas the equations become...

$v=\pi\ r^{2}\ h$ (1)

$a=2\ \pi\ r\ (r+h)$ (2)

... and from them...

$a= 2\ \pi\ r^{2} + \frac{2\ v}{r}$ (3)

Now applying the stand procedure we find the r for minimal area from the equation...

$\frac{d a}{dr}+ 4\ \pi\ r - \frac{2\ v}{r^{2}}=0$ (4)

... the solution of which is...

$r=(\frac{v}{2\ pi})^{\frac{1}{3}}$ (5)

For curiosity I mesured the dimensions of a can of 'Red Kidney-Beans' [my preferred (Nod)...] finding $r=3.75,\ h=11.5\ \implies v=508$. Applying (5) for $v=508$ we obtain an optimum value $r=4.32$... not very far from $r=3.75$ which requires however little more metal (Thinking)...

Kind regards

$\chi$ $\sigma$
• Dec 9th 2011, 09:42 AM
ffezz
Re: Minimization problem
Dont forget that not only are they trying to minimize the metal used in the can, but also trying to optimize shipping and storage as well as production. Perhaps their cans fit better in boxes or something? :D