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To solve the problem, let’s carefully analyze the information provided.

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### Given:
1. The directional derivative of \( f(x, y) \) at \( P \) in the direction of \( \mathbf{v}_1 = 2\mathbf{i} - \mathbf{j} \) is \( \sqrt{5} \).
2. The directional derivative of \( f(x, y) \) at \( P \) in the direction of \( \mathbf{v}_2 = \mathbf{i} + 2\mathbf{j} \) is \( 3\sqrt{5} \).
3. We are asked for the **maximum directional derivative** of \( f(x, y) \) at \( P \).

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### Approach:
1. The directional derivative in the direction of a unit vector \( \mathbf{u} \) is given by:
   \[
   D_{\mathbf{u}} f = \nabla f \cdot \mathbf{u}
   \]
   where \( \nabla f \) is the gradient vector of \( f(x, y) \) and \( \mathbf{u} \) is a unit vector.

2. The magnitude of the maximum directional derivative is equal to the magnitude of the gradient:
   \[
   \text{Maximum directional derivative} = \|\nabla f\|
   \]

3. Use the given information to find \( \nabla f \):
   - \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) are not necessarily unit vectors, so we need their magnitudes:
     \[
     \|\mathbf{v}_1\| = \sqrt{2^2 + (-1)^2} = \sqrt{5}, \quad \|\mathbf{v}_2\| = \sqrt{1^2 + 2^2} = \sqrt{5}
     \]
     The unit vectors in these directions are:
     \[
     \mathbf{u}_1 = \frac{\mathbf{v}_1}{\|\mathbf{v}_1\|} = \frac{2\mathbf{i} - \mathbf{j}}{\sqrt{5}}, \quad \mathbf{u}_2 = \frac{\mathbf{v}_2}{\|\mathbf{v}_2\|} = \frac{\mathbf{i} + 2\mathbf{j}}{\sqrt{5}}
     \]

   - The directional derivatives are given as:
     \[
     D_{\mathbf{u}_1} f = \nabla f \cdot \mathbf{u}_1 = \sqrt{5}, \quad D_{\mathbf{u}_2} f = \nabla f \cdot \mathbf{u}_2 = 3\sqrt{5}
     \]

4. Set up a system of equations using the dot product:
   Let \( \nabla f = \langle a, b \rangle \). Then:
   \[
   \frac{1}{\sqrt{5}}(2a - b) = \sqrt{5} \implies 2a - b = 5 \tag{1}
   \]
   \[
   \frac{1}{\sqrt{5}}(a + 2b) = 3\sqrt{5} \implies a + 2b = 15 \tag{2}
   \]

5. Solve for \( a \) and \( b \):
   From (1): \( b = 2a - 5 \). Substitute into (2):
   \[
   a + 2(2a - 5) = 15 \implies a + 4a - 10 = 15 \implies 5a = 25 \implies a = 5
   \]
   Substituting \( a = 5 \) into \( b = 2a - 5 \):
   \[
   b = 2(5) - 5 = 5
   \]

6. Find \( \|\nabla f\| \):
   \[
   \|\nabla f\| = \sqrt{a^2 + b^2} = \sqrt{5^2 + 5^2} = \sqrt{50} = 5\sqrt{2}
   \]

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### Final Answer:
The maximum directional derivative is:
\[
\boxed{5\sqrt{2}}
\]

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Would you like additional clarification or further steps explained? Here are 5 related questions to deepen your understanding:

1. What does the gradient vector represent geometrically?
2. How can you confirm that \( \mathbf{u}_1 \) and \( \mathbf{u}_2 \) are unit vectors?
3. Why is the magnitude of \( \nabla f \) the maximum directional derivative?
4. How would the solution change if the directional derivatives were given in different directions?
5. Can this method generalize to 3D or higher-dimensional problems?

**Tip:** The maximum directional derivative always occurs in the direction of \( \nabla f \).

At a point P in the xy-plane, the directional derivative of f(x,y) in the direction of v1 = 2i - j is sqrt(5) and the directional derivative of f(x,y) in the direction of v2 = i + 2j is 3sqrt(5). What is the maximum directional derivative of f(x,y) at P?

Solve for the maximum directional derivative of f(x,y) at a point P in the xy-plane, where directional derivatives in two given directions v1 = 2i - j and v2 = i + 2j are provided. Learn to use the gradient vector and dot product to determine the solution, ideal for Multivariable Calculus learners.

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