2. ( 2 points) A box with an open top is to be constructed from a square piece of cardboard with dimensions \( 4 \mathrm{~cm} \times 4 \mathrm{~cm} \). This box is constructed by cutting out a square from each of the four corners and bending up the sides. Find the largest volume that such a box can have.

To maximize the volume of the box constructed from a \(4 \, \text{cm} \times 4 \, \text{cm}\) square piece of cardboard, we can denote the side length of the squares cut from each corner as \(x\). After cutting out the corners and folding up the sides, the dimensions of the box will be: - Length = \(4 - 2x\) - Width = \(4 - 2x\) - Height = \(x\) The volume \(V\) of the box can thus be expressed as: \[ V = (4 - 2x)(4 - 2x)(x) = (4 - 2x)^2 x \] To find the largest volume, we can take the derivative of the volume function with respect to \(x\): 1. First, expand the volume function: \[ V = (16 - 16x + 4x^2)x = 4x^3 - 16x^2 + 16x \] 2. Next, we find the derivative \(V'\): \[ V' = 12x^2 - 32x + 16 \] 3. Setting the derivative equal to zero to find critical points: \[ 12x^2 - 32x + 16 = 0 \] 4. Dividing through by 4 simplifies to: \[ 3x^2 - 8x + 4 = 0 \] 5. Using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 3 \cdot 4}}{2 \cdot 3} = \frac{8 \pm \sqrt{64 - 48}}{6} = \frac{8 \pm \sqrt{16}}{6} = \frac{8 \pm 4}{6} \] This gives us the solutions: \[ x = 2 \quad \text{and} \quad x = \frac{2}{3} \] Since \(x = 2\) would mean no box (height would be zero), we only consider \(x = \frac{2}{3}\). 6. Finally, we plug \(x = \frac{2}{3}\) back into the volume equation to find the maximum volume: \[ V = (4 - 2 \cdot \frac{2}{3})^2 \cdot \frac{2}{3} = \left(\frac{8}{3}\right)^2 \cdot \frac{2}{3} = \frac{64}{9} \cdot \frac{2}{3} = \frac{128}{27} \approx 4.74 \, \text{cm}^3 \] Thus, the largest volume that such a box can have is approximately \(4.74 \, \text{cm}^3\).