# Math Help - optimizing - cylinders

1. ## optimizing - cylinders

A cylindrical can is to hold 500 cm^3 of apple juice. The design must take into account that the height must be between 6 and 15 cm, inclusive. How should the can be constructed so that a minimum amount of material will be used in the construction? (Assume there will be no waste).

Can someone explain to me the steps when you have to find a minimum/maximum for the volume/area/surface area. How do i find the values so that l,w, & h all have the same variable???

2. Originally Posted by Tascja
A cylindrical can is to hold 500 cm^3 of apple juice. The design must take into account that the height must be between 6 and 15 cm, inclusive. How should the can be constructed so that a minimum amount of material will be used in the construction? (Assume there will be no waste).

Can someone explain to me the steps when you have to find a minimum/maximum for the volume/area/surface area. How do i find the values so that l,w, & h all have the same variable???
Hello, Tascja,

1. The main condition contains the value which becomes extreme that means minimal or maximal. Withe your problem it is the surface of the cylindrical can:

$A=2 \cdot \pi \cdot r^2 + 2\cdot \pi \cdot r \cdot h$. That is a function with 2 variables.

2. You know additional conditions: The volume of the can must be a constant and the height of the can doesn't exceed a given value:

$V_{cylindre}=\pi \cdot r^2 \cdot h$. Solve for h and plug in this term into the equation of 1..
$h=\frac{V}{\pi \cdot r^2 }$. You'll get the characteristical function:

3. $A(r)=2 \cdot \pi \cdot r^2 + 2\cdot \pi \cdot r \cdot \frac{V}{\pi \cdot r^2 }=2 \cdot \pi \cdot r^2 + \frac{2V}{ r }$

4. The surface A is extreme if the first derivative of A equals zero:

$A'(r)=4 \cdot \pi \cdot r - \frac{2V}{ r^2 }$. Thus:

$4 \cdot \pi \cdot r - \frac{2V}{ r^2 }=0 \Longleftrightarrow r^3=\frac{2V}{4 \cdot \pi} \Longleftrightarrow r=\sqrt[3]{\frac{2V}{4 \cdot \pi}}$. Plug in the value for V = 500 and you'll get:

$r \approx 4.3 cm$. Now plug in this value into the equation to calculate h:
$h=\frac{500}{\pi \cdot 18.5}\approx 8.6$. That means the diameter and the height of the can are equal, the can has a quadratical shape. The height is in the bound of the given range.

EB