A function that represents the volume of a cardboard box is V(x) = -0.65x^3 + 4x^2 + 3x, where x is the width of the box. Determine the width that will maximize the volume. What are the restrictions on the width?
The answer is 4.45 and domain is 0<x<6.83
V(x) = x(-.65x^2 + 4x + 3)
= -(x+0.68)(x)(x-6.83)
Now you know why the domain is 0<x<6.83
Don't know what tools you have available. This kind of problem is usually done using calculus:
V'(x) = -1.95x^2 + 8x + 3
V'=0 at x=4.45