# Math Help - optimzation problem #9

1. ## optimzation problem #9

A cylindrical-shaped tin can is to have a capacity of 1000 cm^3.
a) find the dimensions of the can that require the minimum amount of tin and assume that no waste material is allowed. The marketing department has specified that the smallest can the market will accept has a diameter of 6 cm and a height of 4 cm.
b) express the answer for part a as a ratio of height to diameter

so far.... I have
(not sure if it is correct)

V=1000cm^3

(0.5d)^2+(0.5h^2)=r^2

V=pir^2h
1000=[pi][(0.5d)^2+(0.5h)^2][h]

2. Originally Posted by skeske1234
A cylindrical-shaped tin can is to have a capacity of 1000 cm^3.
a) find the dimensions of the can that require the minimum amount of tin and assume that no waste material is allowed. The marketing department has specified that the smallest can the market will accept has a diameter of 6 cm and a height of 4 cm.
b) express the answer for part a as a ratio of height to diameter

so far.... I have
(not sure if it is correct)

V=1000cm^3

(0.5d)^2+(0.5h^2)=r^2 this is not a correct relationship ... r is not the distance from one corner of the can to the other.

V=pir^2h
1000=[pi][(0.5d)^2+(0.5h)^2][h]
$1000 = \pi r^2 h$

$h = \frac{1000}{\pi r^2}$

what will be minimized is the surface area ...

$S = 2\pi r^2 + 2\pi rh$

$S = 2\pi r^2 + 2\pi r \cdot \frac{1000}{\pi r^2}$

$S = 2 \pi r^2 + \frac{2000}{r}$

find $\frac{dS}{dr}$ and minimize.

3. Originally Posted by skeeter
$1000 = \pi r^2 h$

$h = \frac{1000}{\pi r^2}$

what will be minimized is the surface area ...

$S = 2\pi r^2 + 2\pi rh$

$S = 2\pi r^2 + 2\pi r \cdot \frac{1000}{\pi r^2}$

$S = 2 \pi r^2 + \frac{2000}{r}$

find $\frac{dS}{dr}$ and minimize.
ok.. question, would the domain of r be..
3<=r<=1000?

4. Originally Posted by skeske1234
A cylindrical-shaped tin can is to have a capacity of 1000 cm^3.
a) find the dimensions of the can that require the minimum amount of tin and assume that no waste material is allowed. The marketing department has specified that the smallest can the market will accept has a diameter of 6 cm and a height of 4 cm.
b) express the answer for part a as a ratio of height to diameter
based on what the marketing department says ...

$r \ge 3$ cm and $h \ge 4$ cm

since $h \ge 4$ ...

$\frac{1000}{\pi r^2} \ge 4$

$\frac{1000}{4\pi} \ge r^2$

$r \le \sqrt{\frac{250}{\pi}}$

so ...

$3 \le r \le \sqrt{\frac{250}{\pi}}$