Maximizing the volume of a cylindrical postal package

[leprofece]

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.

The answer is 2547 cm³

V cilinder = pir²h
and the sum L + L+H = 60
2L + H = 60
solving for H and putting it into the volume i don't get the answer

Yeah I got by h = 60-2L

 

[MarkFL]

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:

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

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

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

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:

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

Can you demonstrate that this is true?

 

[leprofece]

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:

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

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

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

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:

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

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]

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. (Yes)