To minimize the cost of production for the cylindrical can, we need to find the radius $r$ and height $h$ of the cylinder that minimizes the total cost. Let's solve this step by step.

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1. Volume Constraint:

The volume $V$ of the cylinder is given by: $V = \pi r^2 h$ Given that $V = 18\pi$, we can express $h$ in terms of $r$: $h = \frac{18}{r^2}$

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2. Cost Function:

The total cost $C$ is the sum of the costs of the top, bottom, and curved surface area: $C = \text{Cost of top and bottom} + \text{Cost of curved surface}$

• Top and bottom areas (made from material costing Php 1/cm²): $\text{Total area of top and bottom} = 2\pi r^2$ $\text{Cost of top and bottom} = 1 \times 2\pi r^2 = 2\pi r^2$

• Curved surface area (made from material costing Php 3/cm²): $\text{Curved surface area} = 2\pi r h$ $\text{Cost of curved surface} = 3 \times 2\pi r h = 6\pi r h$

Thus, the total cost is: $C = 2\pi r^2 + 6\pi r h$

Substitute $h = \frac{18}{r^2}$ into the cost function: $C = 2\pi r^2 + 6\pi r \left(\frac{18}{r^2}\right)$ Simplify: $C = 2\pi r^2 + \frac{108\pi}{r}$

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3. Minimizing the Cost:

To minimize the cost, take the derivative of $C$ with respect to $r$ and set it to zero: $\frac{dC}{dr} = \frac{d}{dr} \left( 2\pi r^2 + \frac{108\pi}{r} \right)$ $\frac{dC}{dr} = 4\pi r - \frac{108\pi}{r^2}$

Set $\frac{dC}{dr} = 0$: $4\pi r - \frac{108\pi}{r^2} = 0$ $4r = \frac{108}{r^2}$ $4r^3 = 108$ $r^3 = 27$ $r = 3 \, \text{cm}$

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4. Finding $h$:

Using $h = \frac{18}{r^2}$: $h = \frac{18}{3^2} = \frac{18}{9} = 2 \, \text{cm}$

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5. Verification:

The cost function is: $C = 2\pi r^2 + \frac{108\pi}{r}$ Substitute $r = 3$ and $h = 2$: $C = 2\pi (3^2) + \frac{108\pi}{3}$ $C = 2\pi (9) + 36\pi = 18\pi + 36\pi = 54\pi \, \text{Php}$

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Final Answer:

The radius and height of the can that minimize the cost are: $\boxed{r = 3 \, \text{cm}, \, h = 2 \, \text{cm}}$

Would you like further explanation or additional calculations? Here are some related questions:

1. How do you prove the cost is minimized using the second derivative test?
2. What would the cost be for a different volume constraint?
3. How does the cost change if the curved surface material cost changes to Php 4/cm²?
4. Can this problem be solved using numerical methods instead of calculus?
5. What if the volume is given in liters instead of cubic centimeters?

Tip: When dealing with optimization problems, always ensure constraints are correctly applied before solving.