To find the values of \( x \) that make the volume \( V \) greater than 0, we need to analyze the function you provided:
\[ V(x) = 6400x - 320x^2 + 4x^3 \]
This equation can be factored to determine when \( V(x) > 0 \).
Step 1: Find the roots of the equation \( V(x) = 0 \)
To do this, we will set the volume function to zero:
\[ 6400x - 320x^2 + 4x^3 = 0 \]
We can first factor out common terms:
\[ 4x(1600 - 80x + x^2) = 0 \]
This gives us the first root:
\[ x = 0 \]
Next, we need to solve the quadratic \( x^2 - 80x + 1600 = 0 \). To do this, we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = -80, c = 1600 \):
\[ b^2 - 4ac = (-80)^2 - 4(1)(1600) = 6400 - 6400 = 0 \]
Thus, there is one double root:
\[ x = \frac{80}{2} = 40 \]
Step 2: Analyze the roots and intervals
The roots of \( V(x) \) are:
- \( x = 0 \)
- \( x = 40 \)
The volume function is a cubic polynomial, and the leading coefficient (from \( 4x^3 \)) is positive, which means the function will start at 0 when \( x = 0 \), increase, and then decrease toward negative as \( x \) increases beyond the roots.
Thus, we can summarize the intervals as follows:
- For \( x < 0 \), \( V(x) < 0 \) (not valid since \( x \) must be non-negative).
- For \( 0 < x < 40 \), \( V(x) > 0 \).
- At \( x = 40 \), \( V(x) = 0 \).
- For \( x > 40 \), \( V(x) < 0 \).
Conclusion
The values of \( x \) that make \( V > 0 \) are in the interval:
\[ \text{(0, 40)} \]
Answer
Thus, the correct option is:
A. The values of \( x \) that make \( V > 0 \) are the interval \( (0, 40) \).