A box can be formed by cutting a square out of each corner of a piece of cardboard and folding the sides up. If the piece of cardboard is 62 cm by 62 cm and each side of the square that is cut out has length x​ cm, the function that gives the volume of the box is Vequals3844xminus248x squaredplus4 x cubed. Complete parts​ (a) and​ (b) below.

Question

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Part 1
x
x
62
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Part 1
a. Find the values of x that make Vgreater than0. Select the correct choice below and fill in any answer boxes to complete your choice.
A.
The values of x that make Vgreater than0 are the​ point(s)
  
enter your response here.
​(Use a comma to separate answers as​ needed.)
B.
The values of x that make Vgreater than0 are the interval
  
enter your response here.
​(Type your answer in interval​ notation.)
C.
There are no values of x that make Vgreater than0.

Answer 1

To determine the values of \( x \) for which the volume \( V \) of the box is greater than 0, we need to analyze the cubic function given by \( V = 3844x - 248x^2 + 4x^3 \).

Step 1: Find the critical points by solving for when \( V = 0 \).

We need to find the roots of the cubic equation:

\[ 4x^3 - 248x^2 + 3844x = 0 \]

This can be factored as:

\[ 4x(x^2 - 62x + 961) = 0 \]

Next, we can further simplify it by dividing both sides by 4:

\[ x(x^2 - 62x + 961) = 0 \]

Now, we have one root at \( x = 0 \).

Step 2: Solve the quadratic equation \( x^2 - 62x + 961 = 0 \).

Using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1, b = -62, c = 961 \):

\[ x = \frac{62 \pm \sqrt{(-62)^2 - 4 \cdot 1 \cdot 961}}{2 \cdot 1} \] \[ x = \frac{62 \pm \sqrt{3844 - 3844}}{2} \] \[ x = \frac{62 \pm 0}{2} \] \[ x = 31 \]

Step 3: Critical points

The critical points (or roots) we found are:

1. \( x = 0 \)
2. \( x = 31 \)

Step 4: Test intervals between the critical points.

We will test intervals \( (-\infty, 0) \), \( (0, 31) \), and \( (31, +\infty) \):

1. For \( x < 0 \): Choose \( x = -1 \): \[ V = 3844(-1) - 248(-1)^2 + 4(-1)^3 < 0 \]
2. For \( 0 < x < 31 \): Choose \( x = 1 \): \[ V = 3844(1) - 248(1)^2 + 4(1)^3 > 0 \]
3. For \( x > 31 \): Choose \( x = 32 \): \[ V = 3844(32) - 248(32)^2 + 4(32)^3 < 0 \]

Step 5: Analyze the volume sign change

The volume \( V \) is:

• Negative for \( x < 0 \)
• Positive for \( 0 < x < 31 \)
• Negative for \( x > 31 \)

Conclusion

The values of \( x \) that make \( V > 0 \) are: Part (a) A. The value(s) of \( x \) that make \( V > 0 \) is:

• 0 < x < 31 (since we need \( x > 0 \))

Part (b) B. The values of \( x \) that make \( V > 0 \) are the interval: \[ (0, 31) \]

Final Answer

A. The values of \( x \) that make \( V > 0 \): \( 0 < x < 31 \)
B. The values of \( x \) that make \( V > 0 \): \( (0, 31) \)