To find the minimum cost of a rectangular box of volume $200 \, \text{cm}^3$ with given material costs, we can follow these steps:

1. Define variables:

   • Let the dimensions of the box be $x$, $y$, and $z$.
   • Volume constraint: $xyz = 200$.

2. Surface area and cost calculation:

   • The cost of the top and bottom: $2 \cdot (xy + xy) = 4xy$ cents.
   • The cost of the sides: $5 \cdot (2xz + 2yz) = 10xz + 10yz$ cents.

   The total cost $C$ in cents is given by: $C = 4xy + 10xz + 10yz$

3. Volume constraint: $xyz = 200$ Solving for $z$: $z = \frac{200}{xy}$

4. Substitute $z$ into the cost function: $C = 4xy + 10x \left(\frac{200}{xy}\right) + 10y \left(\frac{200}{xy}\right)$ $C = 4xy + \frac{2000}{y} + \frac{2000}{x}$

5. Minimize the cost function:

   • We need to minimize $C = 4xy + \frac{2000}{y} + \frac{2000}{x}$.

   To do this, we can use calculus. First, find the partial derivatives with respect to $x$ and $y$ and set them to zero.

   $\frac{\partial C}{\partial x} = 4y - \frac{2000}{x^2} = 0$ $4y = \frac{2000}{x^2}$ $y = \frac{500}{x^2}$

   $\frac{\partial C}{\partial y} = 4x - \frac{2000}{y^2} = 0$ $4x = \frac{2000}{y^2}$ $x = \frac{500}{y^2}$

6. Solve for $x$ and $y$: $y = \frac{500}{x^2}$

   Substitute $y$ into the volume constraint: $x \left(\frac{500}{x^2}\right) z = 200$ $\frac{500z}{x} = 200$ $z = \frac{200x}{500} = \frac{2x}{5}$

7. Equate $x$ and $y$: Since both equations are symmetrical, $x = y$.

   Substituting $y = \frac{500}{x^2}$ into $x = y$: $x = \frac{500}{x^2}$ $x^3 = 500$ $x = \sqrt[3]{500} \approx 7.937$ Thus, $y \approx 7.937$ $z \approx \frac{2 \cdot 7.937}{5} \approx 3.175$

8. Calculate the minimum cost: $C = 4xy + 10xz + 10yz$ Substituting the values: $C = 4 \cdot 7.937 \cdot 7.937 + 10 \cdot 7.937 \cdot 3.175 + 10 \cdot 7.937 \cdot 3.175$ $C \approx 4 \cdot 63.04 + 10 \cdot 25.2 + 10 \cdot 25.2$ $C \approx 252.16 + 252 + 252$ $C \approx 756.16 \text{ cents}$

Rounding to the nearest whole number: $C \approx 756 \text{ cents}$

Therefore, the minimum cost is approximately 756 cents.

Would you like more details or have any questions? Here are some related questions you might be interested in:

1. How do you find critical points in multivariable functions?
2. What are some applications of Lagrange multipliers?
3. Can you explain the concept of surface area to volume ratio?
4. What is the significance of partial derivatives in optimization problems?
5. How do you use the second derivative test for multivariable functions?

Tip: When solving optimization problems with constraints, it's often helpful to express one variable in terms of the others to reduce the complexity of the function you need to minimize or maximize.