Question #39008

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.

#"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#