1. where the derivative = 0
2. where the derivative does not exist
3. the endpoints of the domain
In your case, .
For #1: Compute (a quadratic polynomial in x), then solve for .
For #2: V(x) is differentiable everywhere, so nothing for #2.
For #3: Need the endpoints of the domain of V. Since x is representing a dimension of a box, must have x>0.
Also, since V represents volume, must have V(x) > 0. This is actually what restricts the domain, and takes some work to solve.
First solve V(x) = 0, the decide on which intervals V is strickly positive. (This approach relies on the Intermediate Value Theorem.)
If and , then
But also, so the solutions to are .
Thus x is either always positive, or always negative, on the intervals .
Either by plugging in a single test value (say x = 7, so computing V(7), or by observing that V's lead coefficent is negative), it's clear the V(x) is negative for . V(1) = -0.65 + 4 + 3 > 6 > 0, so V(x) is positive for . We can stop there, because V's domain requires .
Therefore the domain of V is (0, 6.83).
But as x approaches the endpoints of V's domain, V(x) goes to 0 (Note V(0) = V(6.83) = 0, and V is continuous everywhere).
Thus the endpoints of the domain will not produce anything like a maximum.
Therefore, the maximum will be produced by solving case #1 above.