Question #39008

2 Answers
Aug 22, 2016

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.

enter image source here
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:

enter image source here

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.

Aug 23, 2016

"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