Finding the size of a can

[phawksbball24]

The material for a can is cut from sheets of metal. The cylindrical sides are formed by bending rectangles: these rectangles are cut fromt he sheet with little or no waste. But if the top and bottom discs are cut from squares of side 2r, this leaves considerable waste metal, which may be recycled but has little or no value to the can makers. If this is the case, show that the amount of metal used is minimized when h/r=8/pi or appx 2.55

I think the part that is confusing me the most with this is finding my original equation to get the derivative from, the formula i came up with for finding the surface area is 2pi*rh + 2(4r^2 - pi*r^2) but I am pretty sure this is wrong or I am confused what to do because there is two variables and im not sure how to find the derivative, any help would be appreciated greatly.

 

[galactus]

The total area of the material is:

\(\displaystyle \L\\A=(2r)^{2}+(2r)^{2}+2{\pi}rh=8r^{2}+2{\pi}rh\)

The volume is:

\(\displaystyle \L\\V={\pi}r^{2}h\)

Solve for h:

\(\displaystyle \L\\h=\frac{V}{{\pi}r^{2}}\)

Sub into the A:

\(\displaystyle \L\\8r^{2}+\frac{2V}{r}\)

Differentiate:

\(\displaystyle \L\\\frac{dA}{dr}=16r-\frac{2V}{r^{2}}=\frac{2(8r^{3}-V)}{r^{2}}\)

Set to 0 and solve for r:

\(\displaystyle \L\\\frac{2(8r^{3}-V)}{r^{2}}=0\)

\(\displaystyle \L\\r=\frac{V^{\frac{1}{3}}}{2}\)

This is the only critical point because \(\displaystyle \L\\\frac{d^{2}A}{dr^{2}}>0\)

Therefore, the material is least when \(\displaystyle \L\\r=\frac{V^{\frac{1}{3}}}{2}\)

\(\displaystyle \L\\\frac{r}{h}=\frac{r}{\frac{V}{({\pi}r^{2})}}=\frac{\pi}{V}r^{3}\)

for \(\displaystyle \L\\r=\frac{V^{\frac{1}{3}}}{2}, \;\ \frac{\pi}{V}(\frac{V^{\frac{1}{3}}}{2})^{3}=\frac{\pi}{8}\)

Voila!!

 

[phawksbball24]

thank you very much that helped a lot, and now the 3rd part of the problem is confusing me a little bit. It says, the values of h/r that we found in porblem 1 and 2 (two is using a hexagon to cut the top and bottom portions of the can rather than a square) are a little closer to the ones that actually occur on supermarket shelves, but they still dont account for everything. If we look more closely at some real cans, we see that the lid and the base are formed from discs with a radius larger than r that are bent over the ends of the can. If we allow for this we would increase h/r. More significantly, in addition to the cost of the metalwe need to incorporate the manufacturing of the can into the cost. Lets assume that most of the expense is incurred in joining the sides to the rims of the cans. If we cut the discs from hexagaons as in problem 2, then the total cost is proportional to

4(square root of 3)r^2 + 2pi*rh + k(4pi*r + h)

where k is the reciprocal of the length that can be joined for the cost of one unit area of metal. Show that this expression is minimized when

(cube root or v)/k = (cube root pi*h/r) * (2pi - h/r) / (pi*h/r - 4 (square root 3))

I am fairly confused on this one but im pretty sure that i need to differentiate the equation for total cost first right?

 

[kjvalm]

I am having the same calculus problem

I got lost on this part. Where did you got the r/h? the problem says it is h/r. After that what did you subsitute in?

[FONT=MathJax_Main]\L[/FONT][FONT=MathJax_Math]r[/FONT][FONT=MathJax_Math]h[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math]r[/FONT][FONT=MathJax_Math]V[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]π[/FONT][FONT=MathJax_Math]r[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main] [/FONT][FONT=MathJax_Math]f[/FONT][FONT=MathJax_Math]r[/FONT][FONT=MathJax_Math]a[/FONT][FONT=MathJax_Math]c[/FONT][FONT=MathJax_Math]π[/FONT][FONT=MathJax_Math]V[/FONT][FONT=MathJax_Math]r[/FONT][FONT=MathJax_Main]3[/FONT]​

for

[FONT=MathJax_Main]\L[/FONT][FONT=MathJax_Math]r[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math]V[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Main] [/FONT][FONT=MathJax_Math]π[/FONT][FONT=MathJax_Math]V[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]V[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math]π[/FONT][FONT=MathJax_Main]8[/FONT]

​

 

[HallsofIvy]

The material for a can is cut from sheets of metal. The cylindrical sides are formed by bending rectangles: these rectangles are cut fromt he sheet with little or no waste. But if the top and bottom discs are cut from squares of side 2r, this leaves considerable waste metal, which may be recycled but has little or no value to the can makers. If this is the case, show that the amount of metal used is minimized when h/r=8/pi or appx 2.55

I think the part that is confusing me the most with this is finding my original equation to get the derivative from, the formula i came up with for finding the surface area is 2pi*rh + 2(4r^2 - pi*r^2) but I am pretty sure this is wrong or I am confused what to do because there is two variables and im not sure how to find the derivative, any help would be appreciated greatly.

\(\displaystyle 4r^2- \pi r^2\) is NOT the surface of each end- it is the amount of metal left (wasted) after the circular end is cut out. Now, what is meant by "the amount of metal used"? If it is only the amount of metal in the can itself, that is \(\displaystyle 2\pi rh+ 2\pi r^2\). If it is the metal required, the metal actually used plus the metal wasted, that is \(\displaystyle 2\pi rh+ 4r^2\).