Patricia wants to make a cardboard box with a volume of \( 64 \mathrm{~cm}^{3} \). The cardboard box will be a cube or cuboid, with 6 faces, including a lid. There are many ways to do this, for example: What is the smallest amount of cardboard Patricia could use to construct her box? What length, width and height would this box have?

To find the smallest amount of cardboard Patricia could use to construct her box, we need to determine the dimensions of the cube or cuboid that will have the smallest surface area while still having a volume of \(64 \, \text{cm}^3\). Let's denote the length, width, and height of the cube or cuboid as \(l\), \(w\), and \(h\) respectively. Given: - Volume of the box = \(64 \, \text{cm}^3\) The volume of a cube or cuboid is given by: \[ V = l \times w \times h \] Since the volume is \(64 \, \text{cm}^3\), we have: \[ l \times w \times h = 64 \] To minimize the surface area, we need to find the dimensions that minimize the surface area while satisfying the volume constraint. The surface area of a cube or cuboid is given by: \[ A = 2lw + 2lh + 2wh \] We need to find the dimensions that minimize the surface area while satisfying the volume constraint. Let's denote the surface area as \(A\) and the volume as \(V\). We have the following equations: \[ l \times w \times h = 64 \] \[ 2lw + 2lh + 2wh = A \] We can solve these equations to find the dimensions that minimize the surface area while satisfying the volume constraint. Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}lwh=64\\2lw+2lh+2wh=A\end{array}\right.\) - step1: Calculate: \(\left\{ \begin{array}{l}hlw=64\\2lw+2hl+2hw=A\end{array}\right.\) - step2: Rewrite the expression: \(\left\{ \begin{array}{l}hlw=64\\A=2lw+2hl+2hw\end{array}\right.\) - step3: Substitute the value of \(A:\) \(hlw=64\) - step4: Evaluate: \(lwh=64\) - step5: Divide both sides: \(\frac{lwh}{lw}=\frac{64}{lw}\) - step6: Divide the numbers: \(h=\frac{64}{lw}\) - step7: Substitute the value of \(h:\) \(A=2lw+2\times \frac{64}{lw}\times l+2\times \frac{64}{lw}\times w\) - step8: Simplify: \(A=\frac{2l^{2}w^{2}+128l+128w}{lw}\) - step9: Calculate: \(\left(A,h,l,w\right) = \left(\frac{2l^{2}w^{2}+128l+128w}{lw},\frac{64}{lw},l,w\right),\left(l,w\right) \in \mathbb{R}^{2}\) - step10: Alternative Form: \(\textrm{Infinitely many solutions}\) The system of equations has infinitely many solutions, which means there are multiple ways to construct the box with a volume of \(64 \, \text{cm}^3\) while minimizing the surface area. To find the smallest amount of cardboard Patricia could use to construct her box, we need to consider the cube or cuboid with the smallest surface area. Let's consider the cube or cuboid with the smallest surface area. The cube or cuboid with the smallest surface area will have equal dimensions, which means it will be a cube. For a cube, the volume is given by \(V = l^3\), where \(l\) is the length of the cube. Given that the volume is \(64 \, \text{cm}^3\), we have: \[ l^3 = 64 \] Solving for \(l\), we get: \[ l = \sqrt[3]{64} = 4 \, \text{cm} \] Therefore, the smallest amount of cardboard Patricia could use to construct her box is for a cube with dimensions of \(4 \, \text{cm} \times 4 \, \text{cm} \times 4 \, \text{cm}\).