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We are given that a rubber string of unit length is stretched along the \( z \)-axis, where the map \( x = \phi(z) \) defines the position of the string as:

\[
x = \phi(z) = z + \frac{z^2}{2} - \frac{z^3}{3}
\]

We are asked to find where the stretch maximum occurs along the string.

### Approach:

1. **Stretch Factor**: The stretch factor is proportional to the derivative of \( \phi(z) \), i.e., \( \frac{dx}{dz} = \frac{d\phi}{dz} \), which gives us the local stretching at each point.

2. **Find Maximum**: To find where the stretch is maximized, we will:
   - Compute \( \frac{d\phi}{dz} \).
   - Set the second derivative equal to zero to find critical points.
   - Analyze these points to determine the maximum stretch.

Let's begin by differentiating \( \phi(z) \).

\[
\frac{d\phi}{dz} = 1 + z - z^2
\]

We now find the critical points by setting \( \frac{d^2\phi}{dz^2} = 0 \):

\[
\frac{d^2\phi}{dz^2} = 1 - 2z
\]

Setting this equal to zero:

\[
1 - 2z = 0 \quad \Rightarrow \quad z = \frac{1}{2}
\]

At \( z = \frac{1}{2} \), we compute the value of \( \frac{d\phi}{dz} \) to confirm the maximum stretch:

\[
\frac{d\phi}{dz} \Big|_{z = \frac{1}{2}} = 1 + \frac{1}{2} - \left( \frac{1}{2} \right)^2 = 1 + \frac{1}{2} - \frac{1}{4} = \frac{5}{4}
\]

Thus, the stretch is maximized at \( z = \frac{1}{2} \).

### Conclusion:
The maximum stretch occurs at \( z = \frac{1}{2} \).

Would you like further details or clarifications? Here are 5 related questions you might explore:

1. How does the function \( \phi(z) \) change if different terms are added?
2. What is the physical meaning of the second derivative in this context?
3. Could there be a minimum stretch, and how would you find it?
4. How would you generalize this for a string of arbitrary length?
5. How does this solution change if the initial map involves different powers of \( z \)?

**Tip**: When analyzing maximum or minimum problems, always check the first and second derivatives to verify critical points.

A rubber string of unit length is positioned from 0 to 1 on the z-axis and stretched by the map: x = φ(z) = z + z²/2 - z³/3. Where is the stretch maximum in the string?

Explore how to find the maximum stretch in a rubber string using calculus. This problem uses the function φ(z) = z + z²/2 - z³/3 and involves differentiation and critical points analysis. Suitable for university-level students studying calculus.

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