Answer:
Let x be the width of the compartment,
⇒ The length of the compartment = x + 2
⇒ the depth of the compartment = x -1
Thus, the volume of the compartment, [tex]V(x) = (x+2)x(x-1) = x^3 + x^2 - 2x[/tex]
The volume of the compartment must be 8 cubic meters.
⇒ [tex]x^3 + x^2 - 2x = 8[/tex]
⇒ [tex]x^3 + x^2 - 2x - 8=0[/tex]
⇒ [tex](x-2)(x^2+3x+4)=0[/tex]
If [tex]x-2=0\implies x = 2[/tex] and if [tex]x^2+3x+4=0\implies x = \text{a complex number}[/tex]
But, we can not take width as a complex number.
⇒ Width of the compartment = 2 meter.
Length of the compartment = 2 + 2 = 4 meter.
Depth of the compartment = 2 - 1 = 1 meter.
Here, the function that shows the volume of the compartment is,
[tex]V(x) = x^3 + x^2 - 2x[/tex]
When we lot the graph of that function we found,
[tex]V(x)\rightarrow + \infty[/tex] as [tex]x\rightarrow + \infty[/tex]
But we can not take width as infinite.
Therefore, the maximum value of V(x) will be 8 at x = 2.
