Let's solve the problem **step by step**: **Given:** A cardboard square with side length 4 m. You cut equal squares of side \( x \) from each corner and fold up the sides to make an open box. ### 1. **Define the Variables** - Side of cardboard: \( 4 \) m - Side of each cut square: \( x \) m After cutting and folding: - **Length of box:** \( 4 - 2x \) - **Width of box:** \( 4 - 2x \) - **Height of box:** \( x \) ### 2. **Volume Function** The volume \( V \) of the box is: \[ V = \text{length} \times \text{width} \times \text{height} \] \[ V(x) = (4 - 2x)(4 - 2x)(x) = x(4 - 2x)^2 \] ### 3. **Expand the Function** \[ V(x) = x(4 - 2x)^2 \] \[ = x(16 - 16x + 4x^2) \] \[ = 16x - 16x^2 + 4x^3 \] ### 4. **Find the Maximum Volume (Take the Derivative)** Set \(\frac{dV}{dx} = 0\): \[ \frac{dV}{dx} = 16 - 32x + 12x^2 \] Set to zero: \[ 16 - 32x + 12x^2 = 0 \] Divide by 4: \[ 4 - 8x + 3x^2 = 0 \] ### 5. **Solve the Quadratic Equation** \[ 3x^2 - 8x + 4 = 0 \] Use quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a=3, b=-8, c=4 \): \[ x = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 3 \cdot 4}}{6} \] \[ x = \frac{8 \pm \sqrt{64 - 48}}{6} \] \[ x = \frac{8 \pm \sqrt{16}}{6} \] \[ x = \frac{8 \pm 4}{6} \] So: \[ x_1 = \frac{8+4}{6} = 2 \] \[ x_2 = \frac{8-4}{6} = \frac{4}{6} = \frac{2}{3} \] ### 6. **Check Validity** - If \( x = 2 \), \( 4 - 2x = 0 \) (box has zero base, not valid). - If \( x = \frac{2}{3} \): \[ \text{Length} = 4 - 2x = 4 - 2 \left(\frac{2}{3}\right) = 4 - \frac{4}{3} = \frac{8}{3} \] So the box dimensions are: - **Length:** \( \frac{8}{3} \) m - **Width:** \( \frac{8}{3} \) m - **Height:** \( \frac{2}{3} \) m --- ## **Final Answer** ### **The dimensions of the box with the largest volume are:** \[ \boxed{ \frac{8}{3} \text{ m} \times \frac{8}{3} \text{ m} \times \frac{2}{3} \text{ m} } \]