Finding the max volume of a box with an open lid

[Bg5528]

I need help with this problem please.
Finding the max volume of a box with an open lid.
Using a 20 in by 20 inch sheet of cardboard and cutting out the corners.

Equations i been using:
Volume: Length*Width*height
Area: L*W
Perimeter: 2L+2W

Can someone help me get started

 

[mosenja]

you have the volume equation correct, but your constraint equation should be (surface area) = 20in * 20in = 400 in^2. so what is the surface area of a topless box?

figure that out, then its a straightforward application of the lagrange multiplier technique.

 

[Bg5528]

?? ok I am confused wouldn't that be Area if 20in * 20in?, so the surface area of the sheet of cardboard is 400 in^2?.

Then the surface area of the box with no lid would be: SA= 2(hw)+2(hl)+(lw)?

 

[Bg5528]

This may be irelevant but i went and set the surface area of the sheet to the surface area of the box with no lid.

400=2hw+2hl+lw => 400-wl=h(2w+2l) => (400-wl)/(2w+2l)=h

 

[mosenja]

yep, my interpretation of the problem is that you have 400in^2 of cardboard to make as large of a topless box as you can.

So you want to maximize V(l,h,w) = l*h*w given that S(l,h,w) = 2(hw)+2(hl)+(lw) = 400.

If you know the Lagrange multiplier technique, use that. If not, you will need to use the S equation to solve for one of the variables l,h,w (it doesn't matter which) in terms of the other variables, substitute that into the V equation, and then maximize the resultant function of two variables. Make sense?

 

[Bg5528]

Kinda. So i got to use S(l,h,w)= 2(hw)+2(hl)+(lw)=400 and V=lwh?

 

[mosenja]

Kinda. So i got to use S(l,h,w)= 2(hw)+2(hl)+(lw)=400 and V=lwh?

Indeed. Using what you wrote above, we have:

400=2hw+2hl+lw => 400-wl=h(2w+2l) => (400-wl)/(2w+2l)=h


Hence V = lwh = lw*(400-wl)/(2w+2l).

Now what do you do to find the maximum of V as a function of w and l?

 

[Bg5528]

Ok so say we want to maximize w from that equation:
we would have to use the chain rule as well as the quotient rule?

 

[Office_Shredder]

I don't believe this is set up correctly. When you cut the corners off of a piece of paper/cardboard you can fold up the sides to make a box. You will lose surface area depending on how much stuff you have cut away

Here's a picture I found on google that demonstrates the idea

http://i.ehow.com/images/GlobalPhoto/Articles/4855626/143703-main_Full.jpg You have to make your cuts all squares of the same size in order to make the box actually line up

 

[Bg5528]

Ok i understand that you will loose Surface area when you make cuts. Still confused though

 

[mosenja]

I don't believe this is set up correctly. When you cut the corners off of a piece of paper/cardboard you can fold up the sides to make a box. You will lose surface area depending on how much stuff you have cut away

Here's a picture I found on google that demonstrates the idea

http://i.ehow.com/images/GlobalPhoto/Articles/4855626/143703-main_Full.jpg You have to make your cuts all squares of the same size in order to make the box actually line up

You're right. The way I set up the problem is incorrect. Well, you're kind of right. I don't think the problem is asking for foldable tabs as in the picture.

They are probably asking for something like this: http://imgur.com/EEJy4" (sorry for the crappy paint drawing), where you are to cut out the shaded corners. Sorry for any confusion!

Anyway, you still apply the same technique I outlined above: figure out the constraint equation, solve for one of the variables, substitute that into the volume equation, and maximize the volume.

 

[Bg5528]

ok thanks for the help