Question

Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by cutting congruent squares from the corners and folding up the sides. Then find the volume.

Answer 1

The dimension of the open rectangular box is
`8.216* 4.216* 1.392`.

The volume of the box is 8.217 cubic inches.

Explanation:

Given : The open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by cutting congruent squares from the corners and folding up the sides.

To find : The dimensions and the volume of the open rectangular box ?

Solution :

Let the height be 'x'.

The length of the box is '11-2x'.

The breadth of the box is '7-2x'.

The volume of the box is
`V=l* b* h`

`V=(11-2x)* (7-2x)* x`

`V=4x^3-36x^2+77x`

Derivate w.r.t x,

`V'(x)=4(3x^2)-2(36x)+77`

`V'(x)=12x^2-72x+77`

The critical point when V'(x)=0

`12x^2-72x+77=0`

Solve by quadratic formula,

`x=(18+√(93))/(6),(18-√(93))/(6)`

`x=4.607,1.392`

Derivate again w.r.t x,

`V''(x)=24x-72`

Now,
`V''(4.607)=24(4.607)-72=38.568>0` (+ve)

`V''(1.392)=24(1.392)-72=-38.592<0` (-ve)

So, there is maximum at x=1.392.

The length of the box is
`l=11-2x`

`l=11-2(1.392)=8.216`

The breadth of the box is
`b=7-2x`

`b=7-2(1.392)=4.216`

The height of the box is h=1.392.

The dimension of the open rectangular box is
`8.216* 4.216* 1.392`.

The volume of the box is
`V=l* b* h`

`V=8.216* 4.216* 1.392`

`V=48.217\ in.^3`

The volume of the box is 8.217 cubic inches.