a) Notice that:
1)
[tex]6084x-312x^2+4x^3=4x(1521-78x+x^2)=4x(x-39)^2\text{.}[/tex]
Therefore V(x)=0 at x=0 and it has a double root at x=39.
2)
[tex]\begin{gathered} V(-1)=6084(-1)-312(-1)^2+4(-1)^3, \\ V(-1)=-6084-312-4<0. \end{gathered}[/tex]
Therefore, V(x)<0 when x is in the following interval:
[tex](-\infty,0).[/tex]
3)
[tex]V(1)=6084(1)-312(1)^2+4(1)^3>0.[/tex]
Therefore, V(x)>0 when x is in the following set:
[tex](0,39)\cup(39,\infty).[/tex]
b) Since x is a length, then it must be greater than zero, also 2x must be smaller than 78, therefore the values of x that makes sense in the context are in the interval:
[tex](0,39)\text{.}[/tex]
Answer:
a) Option B) The values of x that makes V>0 are in the set:
[tex](0,39)\cup(39,\infty).[/tex]
b) Option A) The values of x that give squares that can be cut out to construct a box are the interval:
[tex](0,39)\text{.}[/tex]
(0,39).
