Let's solve the problem step by step: **Given:** - The terrarium is a rectangular box with dimensions \( x \) (length), \( y \) (width), and \( z \) (height). - Volume \( V = x \cdot y \cdot z = 3.2 \) ft³. - Costs per square foot: - Walls: $5.00 (4 walls) - Floor: $2.00 - Ceiling: $2.00 ### 1. **Express Surface Area and Cost** - **Walls:** There are 4 walls. Two have area \( xz \), and two have area \( yz \): \[ \text{Wall Area} = 2xz + 2yz \] Cost: \( 5(2xz + 2yz) = 10xz + 10yz \) - **Floor:** Area = \( xy \). Cost: \( 2xy \) - **Ceiling:** Area = \( xy \). Cost: \( 2xy \) - **Total Cost (C):** \[ C = 10xz + 10yz + 2xy + 2xy = 10xz + 10yz + 4xy \] ### 2. **Express \( z \) in terms of \( x \) and \( y \)** Given \( xyz = 3.2 \): \[ z = \frac{3.2}{xy} \] ### 3. **Substitute \( z \) in Cost Function** \[ C = 10x\left(\frac{3.2}{xy}\right) + 10y\left(\frac{3.2}{xy}\right) + 4xy \] \[ C = 10 \cdot \frac{3.2}{y} + 10 \cdot \frac{3.2}{x} + 4xy \] \[ C = \frac{32}{y} + \frac{32}{x} + 4xy \] ### 4. **Minimize \( C \) using symmetry (since floor and ceiling cost the same, and walls have same price)** Assume \( x = y \) (by symmetry, and since cost function is symmetric in \( x \) and \( y \)): Let \( x = y \): From volume constraint: \[ x^2z = 3.2 \implies z = \frac{3.2}{x^2} \] Now substitute \( x = y \) in cost function: \[ C = \frac{32}{x} + \frac{32}{x} + 4x^2 = \frac{64}{x} + 4x^2 \] ### 5. **Minimize \( C \) with respect to \( x \)** Take derivative and set to 0: \[ \frac{dC}{dx} = -\frac{64}{x^2} + 8x \] Set to 0: \[ -\frac{64}{x^2} + 8x = 0 \] \[ 8x = \frac{64}{x^2} \] \[ 8x^3 = 64 \] \[ x^3 = 8 \] \[ x = 2 \] Now, \( y = 2 \), \( z = \frac{3.2}{2^2} = \frac{3.2}{4} = 0.8 \) ### 6. **Calculate Minimum Cost** \[ C = \frac{64}{2} + 4(2^2) = 32 + 16 = 48 \] --- ## **Final answers:** \[ x = 2 \text{ ft} \] \[ y = 2 \text{ ft} \] \[ z = 0.8 \text{ ft} \] \[ \text{The minimum cost of the terrarium is } \$48 \] --- **Fill in the blanks:** - \( x = 2 \) ft - \( y = 2 \) ft - \( z = 0.8 \) ft - The minimum cost = **\$48**