Optimize surface area of a plastic water bottle to find the best shape for achieving minimum casing material used. 400 mL

To optimize the surface area of a plastic water bottle with a fixed volume of 400 mL (or 400 cm³), we need to find the dimensions that minimize the surface area for a given volume. The shape of a water bottle is typically cylindrical, and we'll assume this for our optimization. A cylindrical water bottle has two main components contributing to its surface area: the lateral surface (side) and the two circular ends (top and bottom).

### Step 1: Defining Variables
- Let \( r \) be the radius of the bottle.
- Let \( h \) be the height of the cylindrical part.
- The volume of the cylinder is fixed at 400 cm³.

The formulas for the volume and surface area of a cylinder are:

- **Volume**: \( V = \pi r^2 h \)
- **Surface Area**: \( A = 2\pi r^2 + 2\pi r h \)

We are given that \( V = 400 \) cm³, so:

\[
\pi r^2 h = 400
\]

From this, we can solve for \( h \) in terms of \( r \):

\[
h = \frac{400}{\pi r^2}
\]

### Step 2: Substitute \( h \) into the Surface Area Formula
Next, substitute this expression for \( h \) into the surface area equation to express the surface area \( A \) solely in terms of \( r \):

\[
A = 2\pi r^2 + 2\pi r \left(\frac{400}{\pi r^2}\right)
\]

Simplifying:

\[
A = 2\pi r^2 + \frac{800}{r}
\]

### Step 3: Minimize the Surface Area
To find the value of \( r \) that minimizes the surface area, we take the derivative of \( A \) with respect to \( r \), set it equal to zero, and solve for \( r \).

First, find the derivative of \( A \):

\[
\frac{dA}{dr} = 4\pi r - \frac{800}{r^2}
\]

Set the derivative equal to zero:

\[
4\pi r - \frac{800}{r^2} = 0
\]

Multiply through by \( r^2 \) to clear the fraction:

\[
4\pi r^3 = 800
\]

Solve for \( r \):

\[
r^3 = \frac{800}{4\pi} = \frac{200}{\pi}
\]

\[
r = \left(\frac{200}{\pi}\right)^{\frac{1}{3}}
\]

Approximating this value numerically:

\[
r \approx 3.84 \, \text{cm}
\]

### Step 4: Find the Corresponding Height
Now that we have the optimal radius \( r \approx 3.84 \) cm, we can find the corresponding height using the formula for \( h \):

\[
h = \frac{400}{\pi (3.84)^2}
\]

Approximating the value of \( h \):

\[
h \approx 8.61 \, \text{cm}
\]

### Step 5: Conclusion
The optimal shape for the bottle, which minimizes the surface area for a fixed volume of 400 mL, has a radius of approximately **3.84 cm** and a height of approximately **8.61 cm**. This shape minimizes the amount of plastic used to create the bottle.

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Would you like to see more details about this calculation or have any specific questions?

Here are 5 related questions to explore further:
1. How does the shape change if the volume is increased to 500 mL?
2. What would happen to the surface area if the bottle were rectangular rather than cylindrical?
3. Can you optimize the bottle's shape if it has an additional constraint, like a minimum height?
4. How does the thickness of the plastic affect the overall material used?
5. How would the optimization change if the bottle had a conical or spherical top?

**Tip:** When designing containers, minimizing surface area not only saves material but can also help reduce production costs!

Discover how to minimize the surface area of a cylindrical water bottle with a fixed volume of 400 mL. This step-by-step guide uses key geometry and optimization concepts, along with derivative-based problem-solving, suitable for students in Grades 10-12.

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