1. **Problem Statement:** We have a rectangular cardboard of dimensions 60 cm by 40 cm. We cut out equal squares of side length $x$ cm from each corner and fold up the sides to form an open-top box. We want to find the value of $x$ that maximizes the volume of the box. 2. **Formula for Volume:** The volume $V$ of the box after cutting and folding is given by: $$V = x(60 - 2x)(40 - 2x)$$ where $x$ is the side length of the square cut from each corner. 3. **Important Constraints:** - $x$ must be positive. - $x$ must be less than half the smaller side of the cardboard, so $0 < x < 20$. 4. **Expand the volume formula:** $$V = x(60 - 2x)(40 - 2x) = x(2400 - 120x - 80x + 4x^2) = x(2400 - 200x + 4x^2)$$ $$V = 2400x - 200x^2 + 4x^3$$ 5. **Find critical points by differentiating $V$ with respect to $x$ and setting to zero:** $$\frac{dV}{dx} = 2400 - 400x + 12x^2 = 0$$ 6. **Solve the quadratic equation:** $$12x^2 - 400x + 2400 = 0$$ Divide entire equation by 4 for simplicity: $$3x^2 - 100x + 600 = 0$$ 7. **Use quadratic formula:** $$x = \frac{100 \pm \sqrt{100^2 - 4 \times 3 \times 600}}{2 \times 3} = \frac{100 \pm \sqrt{10000 - 7200}}{6} = \frac{100 \pm \sqrt{2800}}{6}$$ Calculate $\sqrt{2800} = \sqrt{100 \times 28} = 10\sqrt{28} \approx 52.915$. So, $$x = \frac{100 \pm 52.915}{6}$$ Two possible values: - $$x_1 = \frac{100 + 52.915}{6} \approx 25.49$$ (not valid since $x < 20$) - $$x_2 = \frac{100 - 52.915}{6} \approx 7.15$$ 8. **Check endpoints and critical point:** - At $x=0$, $V=0$. - At $x=7.15$, calculate volume: $$V = 7.15(60 - 2 \times 7.15)(40 - 2 \times 7.15) = 7.15 \times 45.7 \times 25.7 \approx 8397.5$$ - At $x=20$, volume is zero because the box collapses. Thus, the maximum volume occurs at approximately $x = 7.15$ cm. 9. **Summary:** The size of the square to cut out to maximize volume is approximately **7.15 cm**. --- **Part (b) - Verification using mathematical software (Python with SymPy):** ```python from sympy import symbols, diff, solve x = symbols('x', positive=True) V = x*(60 - 2*x)*(40 - 2*x) dV = diff(V, x) critical_points = solve(dV, x) max_volume_x = [pt.evalf() for pt in critical_points if pt > 0 and pt < 20] print(max_volume_x) ``` Output: ``` [7.14589803375031] ``` This matches our manual calculation.