Answer:
Let x be the width of the compartment,
Then according to the question,
The length of the compartment = x + 2
And the depth of the compartment = x -1
Thus, the volume of the compartment, ![V = (x+2)x(x-1) = x^3 + x^2 - 2x](https://tex.z-dn.net/?f=V%20%3D%20%28x%2B2%29x%28x-1%29%20%3D%20x%5E3%20%2B%20x%5E2%20-%202x)
But, the volume of the compartment must be 8 cubic meters.
⇒ ![x^3 + x^2 - 2x = 8](https://tex.z-dn.net/?f=x%5E3%20%2B%20x%5E2%20-%202x%20%3D%208)
⇒ ![x^3 + x^2 - 2x - 8=0](https://tex.z-dn.net/?f=x%5E3%20%2B%20x%5E2%20-%202x%20-%208%3D0)
⇒ ![(x-2)(x^2+3x+4)=0](https://tex.z-dn.net/?f=%28x-2%29%28x%5E2%2B3x%2B4%29%3D0)
If
and if ![x^2+3x+4=0\implies x = \text{complex number}](https://tex.z-dn.net/?f=x%5E2%2B3x%2B4%3D0%5Cimplies%20x%20%3D%20%5Ctext%7Bcomplex%20number%7D)
But, width can not be the complex number.
Therefore, width of the compartment = 2 meter.
Length of the compartment = 2 + 2 = 4 meter.
And, Depth of the compartment = 2 - 1 = 1 meter.
Since, the function that shows the volume of the compartment is,
![V(x) = x^3 + x^2 - 2x](https://tex.z-dn.net/?f=V%28x%29%20%3D%20x%5E3%20%2B%20x%5E2%20-%202x)
When we lot the graph of that function we found,
V(x) is maximum for infinite.
But width can not infinite,
Therefore, the maximum value of V(x) will be 8.