A box with no top is to be made from a square piece of cardboard with side length 10 inches by cutting out squares from each corner and folding up the sides. If the side length of each cut square is x, what value of x maximizes the volume of the box? (Provide critical points for maximization.)

To maximize the volume of the box formed from the cardboard, we start by expressing the volume \( V \) in terms of \( x \). The volume of the box can be given by: \[ V(x) = x(10 - 2x)(10 - 2x) \] Here, \( 10 - 2x \) is the new length and width of the box after cutting out squares of side length \( x \) from each corner. Expanding the expression for the volume, we have: \[ V(x) = x(10 - 2x)^2 = x(100 - 40x + 4x^2) = 100x - 40x^2 + 4x^3 \] Now we will find the critical points by taking the derivative of \( V(x) \) and setting it equal to zero: \[ V'(x) = 100 - 80x + 12x^2 \] Setting the derivative to zero and solving for \( x \): \[ 12x^2 - 80x + 100 = 0 \] Dividing the entire equation by 4 gives: \[ 3x^2 - 20x + 25 = 0 \] We can use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ x = \frac{20 \pm \sqrt{(-20)^2 - 4 \cdot 3 \cdot 25}}{2 \cdot 3} = \frac{20 \pm \sqrt{400 - 300}}{6} = \frac{20 \pm \sqrt{100}}{6} = \frac{20 \pm 10}{6} \] Calculating the two possible values: \[ x = \frac{30}{6} = 5 \quad \text{and} \quad x = \frac{10}{6} \approx 1.67 \] Since \( x \) must be less than 5 (as \( 10 - 2x > 0 \)), we discard \( x = 5 \) and keep: \[ x \approx 1.67 \] To confirm that this critical point corresponds to a maximum, we can evaluate the second derivative \( V''(x) \): \[ V''(x) = 24x - 80 \] Evaluating \( V''(x) \) at \( x \approx 1.67 \): \[ V''(1.67) = 24(1.67) - 80 = 40.08 - 80 = -39.92 \] Since \( V''(1.67) < 0 \), this confirms that \( x \approx 1.67 \) yields a local maximum for the volume of the box. Thus, \( x \approx 1.67 \) inches maximizes the volume of the box formed from the cardboard.