Answer:
See below.
Explanation:
The volume of the box is given by the expression.
V(x) = x^3 - 2x^2 - 15x
To find the possible dimensions of the box, we need to solve for the values of x that make the volume positive. A box with negative volume is not physically meaningful.
Setting V(x) > 0, we get.
x^3 - 2x^2 - 15x > 0
Factorizing the left-hand side, we get.
x(x^2 - 2x - 15) > 0
Now, we can find the values of x that make each factor positive.
For x > 0, both factors are positive.
For x^2 - 2x - 15 > 0, we can factor it as (x - 5)(x + 3) > 0. This inequality is true when x < -3 or x > 5.
Therefore, the possible dimensions of the box are.
x > 0 and x < -3, or
x > 0 and x > 5.
However, we need to remember that the dimensions of a physical box must be positive. Therefore, the only valid solution is,
x > 0 and x > 5.
So the possible dimensions of the box are.
Length, width, and height > 5 units.