what is the least amount of material that can be used to make a can with a circular base and closed top with capacity of 6 cubic inches?
can someone help me set this problem up?
Yes,
but there is a difficulty with differentiating the surface area equation,
because it contains "h".
In order for the volume to remain fixed at 6 cubic units,
as the radius changes, the height also changes.
Hence, there are 2 variables "r" and "h".
Hence, skeeter's clue needs to be utilised.
This is because "h" can be written in terms of "r" using the volume formula.
Then you can differentiate with respect to "r".
$\displaystyle 6 = \pi r^2 h$
solving for $\displaystyle h$ ...
$\displaystyle \displaystyle h = \frac{6}{\pi r^2}$
surface area equation ...
$\displaystyle A = 2\pi r^2 + 2\pi r h
$
substitute in the expression for $\displaystyle h$ ...
$\displaystyle \displaystyle A = 2\pi r^2 + 2\pi r \cdot \frac{6}{\pi r^2}$
simplify ...
$\displaystyle \displaystyle A = 2\pi r^2 + \frac{12}{r}$
now, find $\displaystyle \displaystyle \frac{dA}{dr}$ , set it equal to $\displaystyle 0$ and minimize.
That radius will allow you to calculate the corresponding height "h".
Then you can use those two parameter values to calculate the minimum amount of material,
by placing them in the surface area equation.
The differentiation found the value of "r" for which surface area is a minimum.