An open box with a volume of 5000 cm3 is to be constructed by taking a piece of cardboard 30 cm by 60 cm, cutting squares of side length x cm from each corner, and folding up the sides. Show that this can be done in two different ways, and find the exact dimensions of the box in each case.
| 60 | |||
| 30 |
Let x be the height, in cm,
(60 ? 2x) cm
be the width, and the length be
cm.
Thus, the volume of the box is as follows.
| V = 5000 | = | x(60 ? 2x)
|
||
| = | 4x3 ? 180x2 + 1800x |
Additionally, it is clear that we must have
30 ? 2x > 0,
and so
0 < x < .
We must now set the volume equal to 0 and solve for x. We will first factor out a constant.
| 4x3 ? 180x2 + 1800x ? 5000 | = | 0 |
| (x3 ? 45x2 + 450x ? 1250) | = | 0 |
We will now factor
x3 ? 45x2 + 450x ? 1250
in order to find the zeros and continue the factoring process. We complete the following division to identify one of the zeros.
| 5 | 1 | ?45 | 450 | ?1250 | |
| 5 | ?200 | 1250 | |||
| 1 | ?40 | 0 | ?
x = is a zero. |
Thus, we have the following.
| x3 ? 45x2 + 450x ? 1250 | = | 0 | ||||
|
|
= | 0 | ||||
Using the quadratic formula, we find the other zeros.
| x | = |
|
||||||
| = |
|
|||||||
Since
| 40 + 10
|
||
| 2 |
> 15,
it is not a zero, but the other solution found using the quadratic formula is a zero.
Therefore, we have identified the only two zeros. (Enter your answers as a comma-separated list.)
x =
This indicates that there are two solutions to the problem, one for each x-value above.
Using the smaller x-value, state the exact dimensions of the box.
| height |
cm |
|
| length |
cm |
|
| width |
cm |
Using the larger x-value, state the exact dimensions of the box.
| height |
cm |
|
| length |
cm |
|
| width |
cm |