To determine the dimensions of the terrarium that will minimize the total cost while maintaining a volume of [tex]\(80 \, \text{ft}^3\)[/tex], let's break down the requirements and the corresponding math.
### Requirements:
1. Volume Constraint: The terrarium must have a volume of [tex]\(80 \, \text{ft}^3\)[/tex].
2. Cost Estimation:
- Walls: [tex]$4.00$[/tex] per square foot.
- Floor: [tex]$5.00$[/tex] per square foot.
- Ceiling: [tex]$5.00$[/tex] per square foot.
### Steps to Solve:
1. Volume Constraint:
[tex]\[ x \times y \times z = 80 \][/tex]
2. Surface Area Calculation:
- The total area of the walls is the sum of the areas of three pairs of opposing sides:
[tex]\[
\text{Total Wall Area} = 2(xz + yz)
\][/tex]
- The floor and ceiling together:
[tex]\[
\text{Floor and Ceiling Area} = 2(xy)
\][/tex]
3. Total Cost Calculation:
[tex]\[
\text{Total Cost} = \text{Cost of Walls} + \text{Cost of Floor} + \text{Cost of Ceiling}
\][/tex]
Using the given cost per square foot:
[tex]\[
\text{Cost of Walls} = 4 \times 2(xz + yz) = 8(xz + yz)
\][/tex]
[tex]\[
\text{Cost of Floor and Ceiling} = 5 \times 2(xy) = 10(xy)
\][/tex]
Hence, the total cost becomes:
[tex]\[
\text{Total Cost} = 8(xz + yz) + 10(xy)
\][/tex]
### Optimal Dimensions and Minimum Cost
Based on the given and calculated criteria, the dimensions [tex]\(x\)[/tex], [tex]\(y\)[/tex], and [tex]\(z\)[/tex] that minimize the total cost and fulfill the volume constraint can be found:
[tex]\[
x \approx 3.9695 \, \text{ft}
\][/tex]
[tex]\[
y \approx 3.9695 \, \text{ft}
\][/tex]
[tex]\[
z \approx 5.0770 \, \text{ft}
\][/tex]
With these dimensions, the minimum cost of constructing the terrarium is:
[tex]\[
\text{Minimum Cost} \approx \$480.03
\][/tex]
### Final Answer
[tex]\[
\begin{array}{l}
x = 3.9695 \, \text{ft}\\
y = 3.9695 \, \text{ft}\\
z = 5.0770 \, \text{ft}
\end{array}
\][/tex]
[tex]\[
\text{The minimum cost of the terrarium is} \, \$ \, 480.03
\][/tex]
