Question

You are designing a cylindrical package. You can spend $4 on packaging, which costs $0.10 per square cm. You would like to determine the maximum volume that you can contain in a cylinder that costs less than $4.

Answer 1

Answer:

V = 19.42 cubic cm.

Step-by-step explanation:

The area of the cylindrical packege is

$A=2\pi r^{2}+2\pi rh$

and, its volume is

$V = \pi r^{2}h$

where r is the radius and h is the height.

If $4 is the maximun you can spend and, each square cm cost $0.10 then, the maximun amount of square cm that is  available is 40 square cm (by a simple rule of three). So

$40=2\pi r^{2}+2\pi rh$

from this we can get that

$h = \dfrac{40-2\pi r^{2}}{2\pi r}$

substituting this in the volume formula leads us to

$V=\pi r^{2}( \dfrac{40-2\pi r^{2}}{2\pi r})$

$V=( \dfrac{40r-2\pi r^{3}}{2})$

For finding the maximun volume we use derivatives (we know that the maximun volume can be found where its derivative is equal to 0). Since the volume depends of the radius, we have

$\dfrac{dV}{dr}=\dfrac{d}{dr}( \dfrac{40r-2\pi r^{3}}{2})$

$\dfrac{dV}{dr}=( \dfrac{40-6\pi r^{2}}{2})$

For the value of r that maximizes the volume we have that $\dfrac{dV}{dr}=0$. So

$0=( \dfrac{40-6\pi r^{2}}{2})$

and

$r= \sqrt{\dfrac{40}{6\pi}}$.

Now that we have the radius that maximizes the volume, the only thing that is left is to replace it in the volume formula:

$V=( \dfrac{40r-2\pi r^{3}}{2})$

Thus $V=( \dfrac{40\sqrt{\dfrac{40}{6\pi}}-2\pi( \dfrac{40}{6\pi})^{\dfrac{3}{2}}}{2})$

After computing this, we get :

V = 19.42 cubic cm.