Maximizing Volume of Rectangular Prism with 120cm x 80cm Sheet Metal

[Noah1]

My question is that I have to find the dimensions of a rectangular prism (cuboid), where none of the faces are square that will maximise its volume. The sheet metal I have to build it from is 120 cm by 80 cm. I don't even know where to start, I can do it if I'm given a height, width or length but nothing is given.

 

[I like Serena]

My question is that I have to find the dimensions of a rectangular prism (cuboid), where none of the faces are square that will maximise its volume. The sheet metal I have to build it from is 120 cm by 80 cm. I don't even know where to start, I can do it if I'm given a height, width or length but nothing is given.

Hi Noah! Welcome to MHB! ;)

I guess we need to see which rectangles we can slice from the sheet metal that form a cuboid, and then figure out which of those cuboids has the highest volume.
Note that a cuboid has maximum volume given a specific area, if it's a cube. That is, if its faces are all squares.
Suppose we try to slice the sheet metal into 6 squares... can we? (Wondering)

 

[Greg]

It's required that none of the faces are square. Did you have something else in mind?

 

[I like Serena]

It's required that none of the faces are square. Did you have something else in mind?

We get a maximum volume with square faces. Any other configuration will have less volume.
So I'm wondering if it's really a requirement for the faces to be square.
I just didn't ask that question yet.
Perhaps we're only talking about a cuboid for which the faces only don't have to be square?

Either way, it will give us a maximum of $V=4^3\text{ dm}^3=64 \text{ L}$.
After that, if we have to, we can try to find a couple of configurations that do not contain squares.
Such as $l=8\text{ dm}, w=4\text{ dm}, h=\frac{4}3\text{ dm}$, giving $V = 8\cdot 4 \cdot \frac{4}3 = 42.7 \text{ L}$.

 

[Noah1]

We get a maximum volume with square faces. Any other configuration will have less volume.
So I'm wondering if it's really a requirement for the faces to be square.
I just didn't ask that question yet.
Perhaps we're only talking about a cuboid for which the faces only don't have to be square?

Either way, it will give us a maximum of $V=4^3\text{ dm}^3=64 \text{ L}$.
After that, if we have to, we can try to find a couple of configurations that do not contain squares.
Such as $l=8\text{ dm}, w=4\text{ dm}, h=\frac{4}3\text{ dm}$, giving $V = 8\cdot 4 \cdot \frac{4}3 = 42.7 \text{ L}$.

Hi
below the question is a template which basically looks like a net for a rectangular prism which is to be used as a guide. It doesn't mention cutting individual shapes and joining together so I imagine it would be a net cut from the sheet of metal and the remaining metal would be wastage.

 

[I like Serena]

Hi
below the question is a template which basically looks like a net for a rectangular prism which is to be used as a guide. It doesn't mention cutting individual shapes and joining together so I imagine it would be a net cut from the sheet of metal and the remaining metal would be wastage.

Ah okay.
So let's fit the template in the sheet and assign for instance the letters $l,w,h$ to the sides.
Which equations can we get then?

 

[Noah1]

Hi
below the question is a template which basically looks like a net for a rectangular prism which is to be used as a guide. It doesn't mention cutting individual shapes and joining together so I imagine it would be a net cut from the sheet of metal and the remaining metal would be wastage.

The other thing is that the example provided to help which is nothing like this one you need to solve V'(x)=0. It also says the corners of the metal are not to be used so it must be a net rather than individual pieces.

- - - Updated - - -

Ah okay.
So let's fit the template in the sheet and assign for instance the letters $l,w,h$ to the sides.
Which equations can we get then?

V = l x w x h
A = l x w
= 120 x 80
= 9600 cm^3
l x w = 9600 cm^2
l x w x h = V^3

 

[I like Serena]

The other thing is that the example provided to help which is nothing like this one you need to solve V'(x)=0. It also says the corners of the metal are not to be used so it must be a net rather than individual pieces.

- - - Updated - - -V = l x w x h
A = l x w
= 120 x 80
= 9600 cm^3
l x w = 9600 cm^2
l x w x h = V^3

Is the provided template something like:
View attachment 5874

If so, we would have for instance $l + 2h = 80\text{ cm}$...

 

[Noah1]

Is the provided template something like:If so, we would have for instance $l + 2h = 80\text{ cm}$...

Yes that's what the template looks like, we have done another one where the length was twice the width but not one with so many variables

- - - Updated - - -

Yes that's what the template looks like, we have done another one where the length was twice the width but not one with so many variables

2h + 2w = 120 cm

 

[I like Serena]

Yes that's what the template looks like, we have done another one where the length was twice the width but not one with so many variables

2h + 2w = 120 cm

Good! (Nod)

So we have:
$$\begin{cases}l+2h=80 \\ 2h+2w=120\end{cases}
\Rightarrow \begin{cases}l=80-2h \\ w=60-h\end{cases}
$$
And we want to optimize $V=l\times w\times h = (80-2h)(60-h)h$.

Looks like we can take the derivative and set it equal to zero, can't we?

 

[Noah1]

Yes that's what the template looks like, we have done another one where the length was twice the width but not one with so many variables

- - - Updated - - -
2h + 2w = 120 cm

I don't know what to do from here am I supposed to solve it like a simultaneous equation?

 

[Greg]

So we have:
$$\begin{cases}l+2h=80 \\ 2h+2w=120\end{cases}
\Rightarrow \begin{cases}l=80-2h \\ w=60-h\end{cases}
$$
And we want to optimize $V=l\times w\times h = (80-2h)(60-h)h$.

Looks like we can take the derivative and set it equal to zero, can't we?

$$f(h)=(80-2h)(60-h)h=2h^3-200h^2+4800h$$

$$f'(h)=6h^2-400h+4800=0\implies h=\dfrac{100\pm20\sqrt7}{3}$$

$$\Rightarrow h=\dfrac{100-20\sqrt7}{3}\approx15.7$$

$$f''(h)=12h-400,\quad f''(15.7)<0$$

so we have found a maximum ($f(h)$ is concave down at $f(15.7)$).

 

[Noah1]

Yes that's what the template looks like, we have done another one where the length was twice the width but not one with so many variables

- - - Updated - - -
2h + 2w = 120 cm

Okay so when I did it I get
$$\frac{400 \pm \sqrt{-400^2 - 4 \times 6 \times 4800}}{ 2 \times 6} \\
\frac{400 \pm \sqrt{44800}}{ 12} \\
\frac{100 \pm \sqrt{11200}}{3} \\
\frac{100 \pm 40 \sqrt{7}}{3}
$$

where am I going wrong

 

[I like Serena]

$$\frac{400 \pm \sqrt{44800}}{ 12} \\
\frac{100 \pm \sqrt{11200}}{3}
$$

where am I going wrong

Let's do this step a little more careful:
$$\frac{400 \pm \sqrt{44800}}{12} \\
= \frac{4\cdot 100 \pm \sqrt{4 \cdot 11200}}{4\cdot 3} \\
= \frac{4\cdot 100 \pm \sqrt{4} \cdot \sqrt{11200}}{4\cdot 3} \\
= \frac{4\cdot 100 \pm 2 \cdot \sqrt{11200}}{4\cdot 3} \\
$$
See? (Wondering)

 

[Noah1]

Let's do this step a little more careful:
$$\frac{400 \pm \sqrt{44800}}{12} \\
= \frac{4\cdot 100 \pm \sqrt{4 \cdot 11200}}{4\cdot 3} \\
= \frac{4\cdot 100 \pm \sqrt{4} \cdot \sqrt{11200}}{4\cdot 3} \\
= \frac{4\cdot 100 \pm 2 \cdot \sqrt{11200}}{4\cdot 3} \\
$$
See? (Wondering)

Your second step with the 4 are you dividing everything by 4 or why have you put a 4 there?

 

[I like Serena]

Your second step with the 4 are you dividing everything by 4 or why have you put a 4 there?

Let me redo that:
$$\frac{400 \pm \sqrt{44800}}{12} \\
= \frac{400 \pm \sqrt{80^2 \cdot 7}}{12} \\
= \frac{400 \pm 80 \cdot \sqrt{7}}{12} \\
= \frac{4 \cdot 100 \pm 4\cdot 20 \cdot \sqrt{7}}{4\cdot 3} \\
= \frac{\cancel{4}\cdot{100} \pm \cancel{4}\cdot{20} \cdot \sqrt{7}}{\cancel{4}\cdot 3} \\
= \frac{100 \pm 20 \sqrt{7}}{3} \\
$$

 

[Noah1]

Let me redo that:
$$\frac{400 \pm \sqrt{44800}}{12} \\
= \frac{400 \pm \sqrt{80^2 \cdot 7}}{12} \\
= \frac{400 \pm 80 \cdot \sqrt{7}}{12} \\
= \frac{4 \cdot 100 \pm 4\cdot 20 \cdot \sqrt{7}}{4\cdot 3} \\
= \frac{\cancel{4}\cdot{100} \pm \cancel{4}\cdot{20} \cdot \sqrt{7}}{\cancel{4}\cdot 3} \\
= \frac{100 \pm 20 \sqrt{7}}{3} \\
$$

Thank you so much for all your help!

 

[soroban]

My question is that I have to find the dimensions of a rectangular prism (cuboid),
where none of the faces are square that will maximize its volume
The sheet metal I have to build it from is 120 cm by 80 cm.

For a given area of sheet metal, the maximum volume is a cube.
There is no unique answer to your problem.

With an 80 by 120 cm sheet, we can make six 40-by-40 squares,
and form a cube with volume [tex]40^3 = 64,000\text{ cm}^3[/tex]

We can alter the dimensions of the cuboid and obtain a lesser volume.
But there is no single answer.

[tex]\begin{array} {ccc}
\text{Dimensions} && \text{Volume} \\
40\times 40.1 \times 39.9 && 63,\!999.6 \\
40 \times 40.01 \times 39.99 && 63,\!999.996 \\
\vdots && \vdots \end{array}[/tex]

 

[HallsofIvy]

"Taking the derivative and setting it equal to 0" will give the general rule that the maximum volume of a cuboid, for either fixed volume or fixed surface area, is a cube! If we require that none of the faces be a square then there is no solution- we can always make the volume greater by approaching a cube- that is by reducing sides of rectangles that are longer than the other side, extending to those that are shorter.