Hello!
The answer is:
The second option:
[tex]Volumen_{max}=(8-2x)(6-2x)*x[/tex]
Why?
From the statement, we know the dimensions of the box, and the length of the sides to be cut (x).
So,
Working with the length of the box:
Let be 8 the length of the cardboard for the length of the box, so, if we cut out the side of length "x", we have:
[tex]Length=(8-(x+x))=(8-2x)[/tex]
Now,
Working with the width of the box:
Let be 6 the length of the cardboard for the width of the box, so, if we cut out the side of length "x", we have:
[tex]Length=(6-(x+x))=(6-2x)[/tex]
Now that we already know the length and the width of the box, we must remember that the bottom of the box will have the same length "x", so, the greatest possible volume of the cardboard box will be:
[tex]Volumen_{max}=Length*Width*Bottom=(8-2x)(6-2x)*x[/tex]
Have a nice day!
