A zoo supplier is building a glass-walled terrarium whose interior volume is to be $270 \mathrm{ft}^{3}$. Material costs per square foot are estimated as shown below. $\begin{array}{ll}\text { Walls: } & \$ 4.00 \\ \text { Floor: } & \$ 5.00 \\ \text { Ceiling: } & \$ 5.00\end{array}$ What dimensions of the terrarium will minimize the total cost? What is the minimum cost? \[ \begin{array}{l} x=\square \mathrm{ft} \\ y=\square \mathrm{ft} \\ \mathrm{z}=\square \mathrm{ft} \end{array} \] The minimum cost of the terrarium is $\$ \square$.

Step 1 ：Let the dimensions of the terrarium be x, y, and z. The volume of the terrarium is given by the product of the dimensions, \(x*y*z = 270\).

Step 2 ：The cost of the terrarium is given by the sum of the costs of the walls, floor, and ceiling. The walls have a total area of \(2xz + 2yz\), the floor has an area of \(xy\), and the ceiling has an area of \(xy\). Therefore, the total cost is \(4*(2xz + 2yz) + 5*xy + 5*xy = 8xz + 8yz + 10xy\).

Step 3 ：We want to minimize this cost subject to the constraint that \(x*y*z = 270\). This is a constrained optimization problem, which can be solved using the method of Lagrange multipliers.

Step 4 ：The Lagrangian function is \(L(x, y, z, λ) = 8xz + 8yz + 10xy + λ(270 - x*y*z)\). Taking the partial derivatives of L with respect to x, y, z, and λ and setting them equal to zero gives a system of four equations.

Step 5 ：Solving this system will give the values of x, y, z, and λ that minimize the cost. After finding the dimensions that minimize the cost, we can substitute these values into the cost function to find the minimum cost.

Step 6 ：The dimensions that minimize the cost of the terrarium are \(x = \boxed{6} \, \text{ft}\), \(y = \boxed{6} \, \text{ft}\), and \(z = \boxed{\frac{15}{2}} \, \text{ft}\).

Step 7 ：The minimum cost of the terrarium is \$ \boxed{1080}.