# Question #39008

Aug 22, 2016

The dimensions of the box are $11.1 c m \times 52 c m \times 6 c m$, but this box only exists in my head. No such box exists in reality.

#### Explanation:

It always helps to draw a diagram.

Originally, the box had dimensions $l$ (length, which is not known) and $w$ (width, which is also not known). However, when we cut out the squares of length $6$, we get this:

If we were to fold the red areas up to form the sides of the box, the box would have height $6$. The width of the box would be $w - 12 + 6 + 6 = w$, and the length would be $l - 12$. We know $V = l w h$, so:
$V = \left(l - 12\right) \left(w\right) \left(6\right)$

But the problem says the volume is $3456$, so:
$3456 = 6 w \left(l - 12\right)$

Now we have this system:
$1200 = l w \text{ equation 1}$
$3456 = 6 w \left(l - 12\right) \text{ equation 2}$

Solving for $w$ in equation 1, we have:
$w = \frac{1200}{l}$

Plugging this in for $w$ in equation 2, we have:
$3456 = 6 w \left(l - 12\right)$
$3456 = 6 \left(\frac{1200}{l}\right) \left(l - 12\right)$
$3456 = \left(\frac{7200}{l}\right) \left(l - 12\right)$
$3456 = 7200 - \frac{86400}{l}$
$\frac{86400}{l} = 3744$
$86400 = 3744 l \to l \approx 23.1$ cm

We know that $w = \frac{1200}{l}$, and we can use this to solve for the width:
$w = \frac{1200}{23.1} \approx 52$ cm

Note that these are the dimensions on the original metal sheet. When we take out the $6$ cm squares to form the box, the length changes by $12$. Therefore the length of the box is $23.1 - 12 = 11.1$ cm.

When you check the dimensions of $l \times w \times h \to 11.1 c m \times 52 c m \times 6 c m$, you'll see that the volume is off by a little, due to rounding.

Aug 23, 2016

$\text{The volume of the box} = 3456 c {m}^{3}$
$\text{The height of the box} = 6 c m$

$\text{The base area of the box}$
$= \text{Its volume"/"height} = \frac{3456}{6} = 576 c {m}^{2}$

Now let the length of the box be a cm and its width be b cm.
Then $a b = 576. \ldots . \left(1\right)$
To keep the volume and height of the box at given value its base area $a \times b$ must be fixed at $576 c {m}^{2}$

$\text{Now area of its 4 sides}$
$= 2 \left(a + b\right) 6 = 12 \left(a + b\right) c {m}^{2}$

To construct the box 4 squares of dimension $\left(6 \times 6\right) c {m}^{2}$ have been cut off.
So
$a b + 12 \left(a + b\right) + 4 \cdot 6 \cdot 6 = \text{Area of the sheet} \ldots \left(2\right)$

Now let us see what happens if we try to find out a and b using equation (1) and (2).

Combining (1) and (2) we get

$576 + 12 \left(a + b\right) + 144 = \text{sheet area} = 1200$
$\implies 12 \left(a + b\right) = 1200 - 576 - 144 = 480$
$\implies a + b = 40$

Now trying to find out $a - b$
${\left(a - b\right)}^{2} = {\left(a + b\right)}^{2} - 4 a b = {40}^{2} - 4 \cdot 576$
$\implies {\left(a - b\right)}^{2} = 1600 - 2304 < 0$

This shows that real solution is not possible with sheet area 1200cm^2.

But a real solution is possible with a minimum value of the perimeter of the base of the box i.e.$2 \left(a + b\right)$ i.e.$a + b$

$\text{Now } \left(a + b\right) = {\left(\sqrt{a} - \sqrt{b}\right)}^{2} + 2 \sqrt{a b}$
for real values of a and b, $\left(a + b\right)$ will be minimum iff $\left(\sqrt{a} - \sqrt{b}\right) = 0 \implies a = b$ $\textcolor{red}{\text{as "ab="constant}}$

This gives $a \times b = 576 \implies {a}^{2} = 576$
$\implies a = 24 c m$
and $b = 24 c m$

Then by relation (2)
$\text{Sheet area} = a b + 12 \left(a + b\right) + 144$
$= 576 + 12 \cdot \left(24 + 24\right) + 144 = 1296 c {m}^{2}$

Now with this sheet area of $1296 c {m}^{2}$ the problem can be solved.

And the the dimension of the box then will be

$24 c m \times 24 c m \times 6 c m$