Prove that any cylindrical can of volume k cubic units that is to be made using a minimum of material must have the height equal to the diameter.
volume ...
$\displaystyle k = \pi r^2 h$
material is equivalent to the surface area ...
$\displaystyle A = 2\pi r^2 + 2\pi r h$
solve the volume equation for $\displaystyle h$ in terms of $\displaystyle k$ and $\displaystyle r$, then substitute the derived expression for $\displaystyle h$ into the surface area equation ...
find $\displaystyle \frac{dA}{dr}$ and determine the value of $\displaystyle r$ that minimizes $\displaystyle A$ ... finally determine $\displaystyle h$ from your derived expression.
also remember that $\displaystyle k$ is a constant.
So, the given condition is
= k (Eq. 1)
And we need to minimize the surface area given by
(Eq. 2)
Step 1: Solve equation 1 for h in terms of r and substitue into equation 2.
Step 2: Differentiate equation 2 wrt r. Set derivative equal to zero and solve for r.
Step 3: Substitute r back into equation 1 to get h.
You should then see that r and h will be equal.
Good Luck!
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$\displaystyle h = \frac{k}{\pi r^2}$
$\displaystyle A = 2\pi r^2 + 2\pi r h$
substitute $\displaystyle \frac{k}{\pi r^2}$ for $\displaystyle h$ ...
$\displaystyle A = 2\pi r^2 + 2\pi r \cdot \frac{k}{\pi r^2}$
clean up the algebra ...
$\displaystyle A = 2\pi r^2 + \frac{2k}{r}$
$\displaystyle A = 2\left(\pi r^2 + \frac{k}{r}\right)$
try finding $\displaystyle \frac{dA}{dr}$ again.