Question

By cutting away identical squares from each corner of a rectangular piece of cardboard and folding up the resulting flaps, an open box may be made. If the cardboard is 12 in. long and 10 in. wide, find the dimensions of the box that will yield the maximum volume. (Round your answers to two decimal places.)

in (smallest value)
in
in (largest value)

Answer 1

1.81 inch by 8.38 inch by 6.38 inch

Smallest Value=1.81 inch

Largest Value=8.38 inch

Explanation:

The cardboard is 12 in. long and 10 in. wide

Let the length of the square cut off=x

Length of the box=12-2x

Width of the box=10-2x

Height of the box=x

Volume of the box=lwh

V=x(12-2x)(10-2x)

The dimensions of the box that will yield maximum volume occurs at the point where the derivative of V=0.

[TeX] V^{'}=4\,\left( 30 - 22\,x + 3\,x^{2}\right) [/TeX]

Thus:

4(30-22x+3x²)=0

Since 4≠0

3x²-22x+30=0

Solving for x using a calculator gives:

x=1.81 or x=5.52

x cannot be 5.52 inch since the width is 10 inch and removing 2(5.52) from the width gives a negative result.

When x=1.81 inch

Length of the box=12-2x=12-2(1.81)=8.38inch

Width of the box=10-2x=10-2(1.81)=6.38 inch

Therefore, the dimensions at which the Volume is maximum are: 1.81 inch by 8.38 inch by 6.38 inch