Question

The height, width and length of a small box are consecutive integers with the height being the smallest

of the three dimensions. If the length and width are increased by 2 cm each and the height is tripled,
then the volume of the box is increased by 552 cm3
. Use algebraic techniques to find the dimensions of
the original small box.

Answer 1

9514 1404 393

Answer:

4 cm, 5 cm, 6 cm

Explanation:

If x represents the height of the original box, the other two dimensions are ...

width = x+1

length = x+2

and the volume is ...

V = HWL = x(x +1)(x +2) = x³ +3x² +2x

__

After the change in dimensions, the height is 3x, the width is x+3, and the length is x+4. The new volume is ...

V = HWL = 3x(x+3)(x+4) = 3x³ +21x² +36x

The difference in volumes is ...

(3x³ +21x² +36x) -(x³ +3x² +2x) = 552

In standard form, this equation is ...

2x³ +18x² +34x -552 = 0

Factoring out 2 gives ...

x³ +9x² +17x -276 = 0

Possible rational roots are 1, 2, 3, 4, 6, and other divisors of 276. The largest possibility worth considering is less than ∛276 ≈ 6.5. Trial and error, or a graphing calculator can show us that x=4 is the only real solution to this equation.

The dimensions of the original small box are 4 cm × 5 cm × 6 cm.