1. **State the problem:** A company wants to build a square-based container with a lid. The side walls cost 1 dollar per square meter, and the base and lid cost 2 dollars per square meter. The total cost is 100 dollars. We need to find the dimensions of the container that maximize the volume. 2. **Define variables:** Let $x$ be the length of one side of the square base (in meters), and $h$ be the height of the container (in meters). 3. **Write the cost equation:** - Area of the base = $x^2$ - Area of the lid = $x^2$ - Area of the four side walls = $4xh$ Cost = (cost per m² of base and lid) \times (area of base + area of lid) + (cost per m² of side walls) \times (area of side walls) $$\text{Cost} = 2(x^2 + x^2) + 1(4xh) = 4x^2 + 4xh$$ Given total cost is 100: $$4x^2 + 4xh = 100$$ 4. **Express $h$ in terms of $x$:** $$4xh = 100 - 4x^2$$ $$h = \frac{100 - 4x^2}{4x} = \frac{100}{4x} - \frac{4x^2}{4x} = \frac{25}{x} - x$$ 5. **Write the volume function:** $$V = x^2 h = x^2 \left(\frac{25}{x} - x\right) = x^2 \cdot \frac{25}{x} - x^2 \cdot x = 25x - x^3$$ 6. **Maximize the volume:** Take the derivative of $V$ with respect to $x$: $$\frac{dV}{dx} = 25 - 3x^2$$ Set derivative to zero to find critical points: $$25 - 3x^2 = 0$$ $$3x^2 = 25$$ $$x^2 = \frac{25}{3}$$ $$x = \sqrt{\frac{25}{3}} = \frac{5}{\sqrt{3}} = \frac{5\sqrt{3}}{3}$$ 7. **Check the second derivative:** $$\frac{d^2V}{dx^2} = -6x$$ At $x = \frac{5\sqrt{3}}{3} > 0$, second derivative is negative, so this is a maximum. 8. **Find $h$ at this $x$:** $$h = \frac{25}{x} - x = \frac{25}{\frac{5\sqrt{3}}{3}} - \frac{5\sqrt{3}}{3} = 25 \cdot \frac{3}{5\sqrt{3}} - \frac{5\sqrt{3}}{3} = \frac{75}{5\sqrt{3}} - \frac{5\sqrt{3}}{3} = \frac{15}{\sqrt{3}} - \frac{5\sqrt{3}}{3}$$ Rationalize the first term: $$\frac{15}{\sqrt{3}} = 15 \cdot \frac{\sqrt{3}}{3} = 5\sqrt{3}$$ So, $$h = 5\sqrt{3} - \frac{5\sqrt{3}}{3} = 5\sqrt{3} \left(1 - \frac{1}{3}\right) = 5\sqrt{3} \cdot \frac{2}{3} = \frac{10\sqrt{3}}{3}$$ **Final answer:** The container with maximum volume has base side length $$x = \frac{5\sqrt{3}}{3} \approx 2.89 \text{ meters}$$ and height $$h = \frac{10\sqrt{3}}{3} \approx 5.77 \text{ meters}$$