1. **Problem Statement:** We need to design a square-based, open-top tank with volume 500 ft³ that minimizes the weight of the steel used. 2. **Given:** - Base side length = $x$ ft - Height (depth) = $y$ ft - Volume constraint: $V = x^2 y = 500$ - Surface area (steel area) to minimize: $S = x^2 + 4xy$ - Weight $w = tS$ where $t$ is thickness (constant) 3. **Express height $y$ in terms of $x$ using volume:** $$y = \frac{500}{x^2}$$ 4. **Rewrite surface area $S$ in terms of $x$ only:** $$S = x^2 + 4x \cdot \frac{500}{x^2} = x^2 + \frac{2000}{x}$$ 5. **Weight function:** $$w(x) = t \left(x^2 + \frac{2000}{x}\right)$$ 6. **Find critical points by differentiating $w(x)$ with respect to $x$:** $$w'(x) = t \left(2x - \frac{2000}{x^2}\right)$$ 7. **Set derivative to zero to find critical values:** $$2x - \frac{2000}{x^2} = 0 \implies 2x = \frac{2000}{x^2} \implies 2x^3 = 2000 \implies x^3 = 1000 \implies x = 10$$ 8. **Check second derivative to confirm minimum:** $$w''(x) = t \left(2 + \frac{4000}{x^3}\right)$$ At $x=10$: $$w''(10) = t \left(2 + \frac{4000}{1000}\right) = t(2 + 4) = 6t > 0$$ Since $w''(10) > 0$, $x=10$ is a minimum. 9. **Find corresponding height $y$:** $$y = \frac{500}{10^2} = \frac{500}{100} = 5$$ **Final answer:** The tank dimensions that minimize weight are base edges of 10 ft and height 5 ft.