# Finding the max volume of a box with an open lid

• Bg5528
In summary, someone needs to help me with a problem involving finding the maximum volume of a box with an open lid. They have the volume equation correct, but their constraint equation should be (surface area) = 20in * 20in = 400 in^2. so what is the surface area of a topless box?
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

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.

?? 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)?

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

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?

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

Bg5528 said:
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?

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?

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

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

Office_Shredder said:
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.

Last edited by a moderator:

ok thanks for the help

## 1. How do you find the max volume of a box with an open lid?

To find the max volume of a box with an open lid, you will need to use the formula V = lwh, where V is the volume, l is the length, w is the width, and h is the height of the box. However, since the box has an open lid, the height will be limited by the height of the box itself.

## 2. Can you explain the concept of max volume in more detail?

Max volume refers to the largest possible volume that a box with an open lid can hold. This means that the box is filled to its maximum capacity without any overflow. In other words, it is the point at which the volume cannot increase any further without changing the dimensions of the box.

## 3. What are the variables that affect the max volume of a box with an open lid?

The variables that affect the max volume of a box with an open lid are the length, width, and height of the box. These dimensions determine the overall size and shape of the box, which in turn affects its volume. Additionally, the material used to construct the box can also impact its max volume.

## 4. Is there a specific method for finding the max volume of a box with an open lid?

Yes, there is a specific method for finding the max volume of a box with an open lid. This method involves calculating the volume using the formula V = lwh and then determining the maximum value by considering the constraints of the open lid. In some cases, it may be necessary to use calculus to find the exact max volume.

## 5. Why is it important to find the max volume of a box with an open lid?

Finding the max volume of a box with an open lid is important for several reasons. For one, it allows us to determine the optimal size and shape of a box for a given purpose. It also helps us understand the limitations of a box with an open lid, which can be useful for practical applications such as packaging and storage. Additionally, finding the max volume can also aid in maximizing efficiency and minimizing waste.

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