# [SOLVED]Maximizing the volume of a cylindrical postal package

#### leprofece

##### Member
The sum of the length and the perimeter of base of a postal package to is 60 cm. find the maximum volume:
when the package is cylindrical.

V cilinder = pir2h
and the sum L + L+H = 60
2L + H = 60
solving for H and putting it into the volume i dont get the answer

Yeah I got by h = 60-2L

Last edited:

#### MarkFL

Staff member
Re: max and min 297

Let's let $P$ be the sum of the length and the perimeter of the base. For a cylinder, we then have the constraint:

$$\displaystyle 2\pi r+h=P$$

Now, the volume of the cylinder, our objective function, is:

$$\displaystyle V(r,h)=\pi r^2h$$

Solve the constraint for $h$, then substitute into the objective function for $h$, and you will then have a function in one variable, $r$. At this point you can maximize the function. You should show that the critical value is at a maximum. You should be able to show that:

$$\displaystyle V_{\max}=\frac{P^3}{27\pi}$$

Can you demonstrate that this is true?

#### leprofece

##### Member
Re: max and min 297

Let's let $P$ be the sum of the length and the perimeter of the base. For a cylinder, we then have the constraint:

$$\displaystyle 2\pi r+h=P$$

Now, the volume of the cylinder, our objective function, is:

$$\displaystyle V(r,h)=\pi r^2h$$

Solve the constraint for $h$, then substitute into the objective function for $h$, and you will then have a function in one variable, $r$. At this point you can maximize the function. You should show that the critical value is at a maximum. You should be able to show that:

$$\displaystyle V_{\max}=\frac{P^3}{27\pi}$$

Can you demonstrate that this is true?
my solving is
pir2(60-2pir)
v= 60pir2-2pi2r2
V
dv = 120 pir -6pi2r
solving r= 20/pi
and h = 20
v= 8000/pi
Then v = 2547 cm3

#### MarkFL

Staff member
Re: max and min 297

my solving is
pir2(60-2pir)
v= 60pir2-2pi2r2
V
dv = 120 pir -6pi2r
solving r= 20/pi
and h = 20
v= 8000/pi
Then v = 2547 cm3
Yes, 8000/pi is the exact answer.