To solve this problem step by step, let's break it down into parts.
Part 1
(a) The height of the box is equal to the side length of the squares cut out, which is \( x \). Therefore, the height of the box is \[ \text{Height} = x \text{ inches}. \]
Part 2
(b) The dimensions of the base can be found by subtracting twice the length of the square from the original dimensions of the piece of tin. Since the original dimensions are 24 inches by 24 inches:
- The length of the base \( l \) will be \( 24 - 2x \).
- The width of the base \( w \) will also be \( 24 - 2x \).
Thus, the dimensions of the base of the constructed box are \[ (24 - 2x) \text{ inches by } (24 - 2x) \text{ inches}. \]
Part 3
(c) Using the formula \( V = lwh \), we can substitute the values we found: \[ V = (24 - 2x)(24 - 2x)(x). \] This simplifies to: \[ V = (24 - 2x)^2 \cdot x. \] Let's expand this: \[ V = (24 - 2x)(24 - 2x) \cdot x = (576 - 96x + 4x^2)x = 4x^3 - 96x^2 + 576x. \]
Part 4
(d) We need to find the values of \( x \) that make \( V = 0 \): \[ 4x^3 - 96x^2 + 576x = 0. \] Factoring out \( 4x \): \[ 4x(x^2 - 24x + 144) = 0. \] So, \[ 4x = 0 \quad \text{or} \quad x^2 - 24x + 144 = 0. \] The first equation gives us \( x = 0 \).
Now, we solve the quadratic equation: \[ x^2 - 24x + 144 = 0. \] Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ x = \frac{24 \pm \sqrt{(-24)^2 - 4 \cdot 1 \cdot 144}}{2 \cdot 1} = \frac{24 \pm \sqrt{576 - 576}}{2} = \frac{24 \pm 0}{2} = 12. \] The solutions are \( x = 0 \) and \( x = 12 \). Thus, \( x \) values that make \( V = 0 \) are: \[ x = 0, 12. \]
Part 5
(e) We need to determine for which of these values of \( x \) a box can exist:
- When \( x = 0 \), the box cannot be formed; it would just be a flat piece of tin.
- When \( x = 12\), cutting a square of size 12 inches out of a 24-inch piece of tin results in a base of \( 0 \) (since \( 24 - 2 \times 12 = 0 \)), hence a box cannot exist either.
Therefore, the correct answer is: D. For neither of the values of \( x \).