A point $P$ is given on the curve $x^4+y^4=1$. Find the maximum distance from the point $P$ to the origin.
1. We start with the given curve equation:
   \[
   x^4 + y^4 = 1
   \]
   We need to find the maximum distance from any point \( P = (x, y) \) on this curve to the origin \( O = (0, 0) \).

2. The distance \( d \) from the point \( P \) to the origin is given by:
   \[
   d = \sqrt{x^2 + y^2}
   \]

3. To find the maximum distance, we need to maximize \( \sqrt{x^2 + y^2} \) subject to the constraint \( x^4 + y^4 = 1 \).

4. We use the method of Lagrange multipliers. Define the function:
   \[
   f(x, y) = x^2 + y^2
   \]
   and the constraint:
   \[
   g(x, y) = x^4 + y^4 - 1 = 0
   \]

5. The Lagrange function is:
   \[
   \mathcal{L}(x, y, \lambda) = x^2 + y^2 + \lambda (x^4 + y^4 - 1)
   \]

6. Taking partial derivatives and setting them to zero:
   \[
   \frac{\partial \mathcal{L}}{\partial x} = 2x + 4\lambda x^3 = 0
   \]
   \[
   \frac{\partial \mathcal{L}}{\partial y} = 2y + 4\lambda y^3 = 0
   \]
   \[
   \frac{\partial \mathcal{L}}{\partial \lambda} = x^4 + y^4 - 1 = 0
   \]

7. From the first two equations:
   \[
   2x(1 + 2\lambda x^2) = 0 \quad \Rightarrow \quad x = 0 \text{ or } 1 + 2\lambda x^2 = 0
   \]
   \[
   2y(1 + 2\lambda y^2) = 0 \quad \Rightarrow \quad y = 0 \text{ or } 1 + 2\lambda y^2 = 0
   \]

8. If \( x = 0 \), then \( y^4 = 1 \) which gives \( y = \pm 1 \). Similarly, if \( y = 0 \), then \( x^4 = 1 \) which gives \( x = \pm 1 \).

9. If \( 1 + 2\lambda x^2 = 0 \), then \( \lambda = -\frac{1}{2x^2} \). Similarly, \( \lambda = -\frac{1}{2y^2} \). Equating these:
   \[
   -\frac{1}{2x^2} = -\frac{1}{2y^2} \quad \Rightarrow \quad x^2 = y^2
   \]

10. Substituting \( x^2 = y^2 \) into the constraint \( x^4 + y^4 = 1 \):
    \[
    2x^4 = 1 \quad \Rightarrow \quad x^4 = \frac{1}{2} \quad \Rightarrow \quad x^2 = \frac{1}{\sqrt{2}} \quad \Rightarrow \quad x = \pm \sqrt[4]{2}
    \]
    Similarly, \( y = \pm \sqrt[4]{2} \).

11. Therefore, the points on the curve that maximize the distance are \( (\sqrt[4]{2}, \sqrt[4]{2}) \) and \( (-\sqrt[4]{2}, -\sqrt[4]{2}) \).

12. The maximum distance is:
    \[
    d = \sqrt{(\sqrt[4]{2})^2 + (\sqrt[4]{2})^2} = \sqrt{\sqrt{2} + \sqrt{2}} = \sqrt{2\sqrt{2}} = \sqrt{\sqrt{4}} = \sqrt{2}
    \]

The final answer is \(\boxed{\sqrt{2}}\).