# Math Help - - Optimization Problems - Proving Question

1. ## - Optimization Problems - Proving Question

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.

2. Originally Posted by darkknight100
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 ...

$k = \pi r^2 h$

material is equivalent to the surface area ...

$A = 2\pi r^2 + 2\pi r h$

solve the volume equation for $h$ in terms of $k$ and $r$, then substitute the derived expression for $h$ into the surface area equation ...

find $\frac{dA}{dr}$ and determine the value of $r$ that minimizes $A$ ... finally determine $h$ from your derived expression.

also remember that $k$ is a constant.

3. 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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4. Originally Posted by dats13
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 h = 2r.
...

5. can you guys actually show the work?

6. Originally Posted by darkknight100
can you guys actually show the work?
you've been given the setup and directions ... follow through with it, please.

7. volume equation for in terms of and , then substitute the derived expression for

what do you mean by that?

8. Originally Posted by darkknight100
volume equation for in terms of and , then substitute the derived expression for

what do you mean by that?
this ...

$k = \pi r^2 h$

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

9. Originally Posted by skeeter
...
Yes, h=2r. Sorry, my mistake.

10. how do you solve for r for the following

4pi r^3 + 2r - 2k
______________ = 0
r^2

11. Originally Posted by darkknight100
how do you solve for r for the following

4pi r^3 + 2r - 2k
______________ = 0
r^2
how did you get that equation?

12. derivative of the surface area equation

13. $h = \frac{k}{\pi r^2}$

$A = 2\pi r^2 + 2\pi r h$

substitute $\frac{k}{\pi r^2}$ for $h$ ...

$A = 2\pi r^2 + 2\pi r \cdot \frac{k}{\pi r^2}$

clean up the algebra ...

$A = 2\pi r^2 + \frac{2k}{r}$

$A = 2\left(\pi r^2 + \frac{k}{r}\right)$

try finding $\frac{dA}{dr}$ again.

14. what's the derivative of
k ?
r

15. Originally Posted by darkknight100
what's the derivative of
k ?
r
what's the derivative of $kr^{-1}$ ?

(remember that k is a constant)

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