The top and bottom of a right cylindrical can are to be made from a material that costs Php 1 /cm2 . Its curved surface is to be constructed from another material that costs Php 3 /cm2 . If the can is to enclose a volume of 18π cm3 , find the radius and height of the can that will minimize the cost of production.

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.

---

### 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 \( r \):
\[
h = \frac{18}{r^2}
\]

---

### 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:
\[
C = 2\pi r^2 + \frac{108\pi}{r}
\]

---

### 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 \(\frac{dC}{dr} = 0\):
\[
4\pi r - \frac{108\pi}{r^2} = 0
\]
\[
4r = \frac{108}{r^2}
\]
\[
4r^3 = 108
\]
\[
r^3 = 27
\]
\[
r = 3 \, \text{cm}
\]

---

### 4. **Finding \( h \):**
Using \( h = \frac{18}{r^2} \):
\[
h = \frac{18}{3^2} = \frac{18}{9} = 2 \, \text{cm}
\]

---

### 5. **Verification:**
The cost function is:
\[
C = 2\pi r^2 + \frac{108\pi}{r}
\]
Substitute \( r = 3 \) and \( h = 2 \):
\[
C = 2\pi (3^2) + \frac{108\pi}{3}
\]
\[
C = 2\pi (9) + 36\pi = 18\pi + 36\pi = 54\pi \, \text{Php}
\]

---

### 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.

The top and bottom of a right cylindrical can are to be made from a material that costs Php 1 /cm2. Its curved surface is to be constructed from another material that costs Php 3 /cm2. If the can is to enclose a volume of 18π cm3, find the radius and height of the can that will minimize the cost of production.

Explore how to minimize the production cost of a cylindrical can with a volume of 18π cm³. This solution includes step-by-step calculations using calculus and geometry to find the optimal radius and height. Suitable for high school students in Grades 11-12.

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