1. **State the problem:** We want to find the dimensions of an open-top rectangular box with volume 32 ft³ that uses the least amount of material (surface area). 2. **Define variables:** Let the length be $x$, width be $y$, and height be $z$. 3. **Write the constraints and objective:** - Volume constraint: $$xyz = 32$$ - Surface area to minimize (open top): $$S = xy + 2xz + 2yz$$ 4. **Set up Lagrange multipliers:** We want to minimize $S$ subject to $g(x,y,z) = xyz - 32 = 0$. 5. **Form the Lagrangian:** $$L = xy + 2xz + 2yz - \lambda (xyz - 32)$$ 6. **Find partial derivatives and set to zero:** $$\frac{\partial L}{\partial x} = y + 2z - \lambda yz = 0$$ $$\frac{\partial L}{\partial y} = x + 2z - \lambda xz = 0$$ $$\frac{\partial L}{\partial z} = 2x + 2y - \lambda xy = 0$$ $$\frac{\partial L}{\partial \lambda} = xyz - 32 = 0$$ 7. **From the first three equations:** - From $\frac{\partial L}{\partial x}$: $y + 2z = \lambda yz$ - From $\frac{\partial L}{\partial y}$: $x + 2z = \lambda xz$ - From $\frac{\partial L}{\partial z}$: $2x + 2y = \lambda xy$ 8. **Divide first two equations to find relation between $x$ and $y$:** $$\frac{y + 2z}{x + 2z} = \frac{\lambda yz}{\lambda xz} = \frac{y}{x}$$ Cross-multiplied: $$(y + 2z) x = (x + 2z) y$$ $$xy + 2xz = xy + 2yz$$ Simplify: $$2xz = 2yz \implies x = y$$ 9. **Use $x = y$ in the third equation:** $$2x + 2x = \lambda x x \implies 4x = \lambda x^2 \implies \lambda = \frac{4}{x}$$ 10. **Use $x = y$ and $\lambda = \frac{4}{x}$ in the first equation:** $$y + 2z = \lambda y z \implies x + 2z = \frac{4}{x} x z = 4z$$ $$x + 2z = 4z \implies x = 2z$$ 11. **Now we have:** $$x = y$$ $$x = 2z$$ 12. **Use volume constraint:** $$xyz = 32 \implies x \cdot x \cdot z = 32 \implies x^2 z = 32$$ Substitute $z = \frac{x}{2}$: $$x^2 \cdot \frac{x}{2} = 32 \implies \frac{x^3}{2} = 32 \implies x^3 = 64 \implies x = 4$$ 13. **Find $y$ and $z$:** $$y = x = 4$$ $$z = \frac{x}{2} = 2$$ 14. **Final answer:** The dimensions that minimize material are: $$\boxed{x = 4, y = 4, z = 2}$$