Question

Question #39008

Answer 1

The dimensions of the box are `11.1 cm xx52cmxx6cm`, 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=lwh`, so:
`V=(l-12)(w)(6)`

But the problem says the volume is `3456`, so:
`3456=6w(l-12)`

Now we have this system:
`1200=lw" equation 1"`
`3456=6w(l-12)" equation 2"`

Solving for `w` in equation 1, we have:
`w=1200/l`

Plugging this in for `w` in equation 2, we have:
`3456=6w(l-12)`
`3456=6(1200/l)(l-12)`
`3456=(7200/l)(l-12)`
`3456=7200-86400/l`
`86400/l=3744`
`86400=3744l->l~~23.1` cm

We know that `w=1200/l`, and we can use this to solve for the width:
`w=1200/23.1~~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 `lxxwxxh->11.1cmxx52cmxx6cm`, you'll see that the volume is off by a little, due to rounding.

Answer 2

`"The volume of the box"=3456cm^3`
`"The height of the box"=6cm`

`"The base area of the box"`
`="Its volume"/"height"=3456/6=576cm^2`

Now let the length of the box be a cm and its width be b cm.
Then `ab=576.....(1)`
To keep the volume and height of the box at given value its base area `axxb` must be fixed at `576cm^2`

`"Now area of its 4 sides"`
`=2(a+b)6=12(a+b)cm^2`

To construct the box 4 squares of dimension `(6xx6)cm^2` have been cut off.
So
`ab+12(a+b)+4*6*6="Area of the sheet"...(2)`

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(a+b)+144= "sheet area"=1200`
`=>12(a+b)=1200-576-144=480`
`=>a+b=40`

Now trying to find out `a-b`
`(a-b)^2=(a+b)^2-4ab=40^2-4*576`
`=>(a-b)^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(a+b)` i.e.`a+b`

`"Now "(a+b)=(sqrta-sqrtb)^2+2sqrt(ab)`
for real values of a and b, `(a+b)` will be minimum iff `(sqrta-sqrtb)=0=>a=b` `color(red)("as "ab="constant")`

This gives `axxb=576=>a^2=576`
`=>a=24cm`
and `b=24cm`

Then by relation (2)
`"Sheet area"=ab+12(a+b)+144`
`=576+12*(24+24)+144=1296cm^2`

Now with this sheet area of `1296cm^2` the problem can be solved.

And the the dimension of the box then will be

`24cmxx24cmxx6cm`