Largest Lidless Box Eaf52A

1. **State the problem:** We want to find the dimensions of a lidless rectangular box with the largest volume, given that the surface area of the cardboard used is 4 m^2. 2. **Define variables:** Let the length be $x$, the width be $y$, and the height be $h$. 3. **Surface area constraint:** Since the box is lidless, the surface area is the sum of the base and the four sides: $$S = xy + 2xh + 2yh = 4$$ 4. **Volume to maximize:** The volume $V$ of the box is: $$V = xyz = xyh$$ 5. **Express $h$ in terms of $x$ and $y$ using the surface area constraint:** $$xy + 2xh + 2yh = 4 \implies 2h(x + y) = 4 - xy \implies h = \frac{4 - xy}{2(x + y)}$$ 6. **Substitute $h$ into the volume formula:** $$V = xy \times \frac{4 - xy}{2(x + y)} = \frac{xy(4 - xy)}{2(x + y)}$$ 7. **To find the maximum volume, use symmetry and set $x = y$ (since the problem is symmetric in $x$ and $y$):** Let $x = y$, then: $$h = \frac{4 - x^2}{4x}$$ Volume becomes: $$V = x^2 \times h = x^2 \times \frac{4 - x^2}{4x} = \frac{x(4 - x^2)}{4} = \frac{4x - x^3}{4}$$ 8. **Find critical points by differentiating $V$ with respect to $x$ and setting to zero:** $$\frac{dV}{dx} = \frac{4 - 3x^2}{4} = 0 \implies 4 - 3x^2 = 0 \implies 3x^2 = 4 \implies x^2 = \frac{4}{3} \implies x = \frac{2}{\sqrt{3}}$$ 9. **Calculate $h$ at this $x$:** $$h = \frac{4 - x^2}{4x} = \frac{4 - \frac{4}{3}}{4 \times \frac{2}{\sqrt{3}}} = \frac{\frac{8}{3}}{\frac{8}{\sqrt{3}}} = \frac{8}{3} \times \frac{\sqrt{3}}{8} = \frac{\sqrt{3}}{3}$$ 10. **Final dimensions:** $$x = y = \frac{2}{\sqrt{3}}, \quad h = \frac{\sqrt{3}}{3}$$ These dimensions give the largest volume for the lidless box with surface area 4 m^2.