Find the critical point(s) (x0,y0) of the function f(x,y)=√ (5x^3+2y^3) +2. Enter your answer as an ordered pair and separate multiple critical points with commas. If there are no critical points, enter ∅.

To find the critical points of the function \( f(x, y) = \sqrt{5x^3 + 2y^3} + 2 \), we need to identify the points where the partial derivatives of \( f \) with respect to \( x \) and \( y \) are both zero or do not exist.

### Step 1: Define and Differentiate the Function
The function is:
\[
f(x, y) = \sqrt{5x^3 + 2y^3} + 2
\]

To find the critical points, we calculate the partial derivatives \( f_x \) and \( f_y \) and set them equal to zero.

### Step 2: Compute the Partial Derivatives
1. **Partial derivative with respect to \( x \):**
   \[
   f_x = \frac{d}{dx} \left( \sqrt{5x^3 + 2y^3} + 2 \right) = \frac{1}{2\sqrt{5x^3 + 2y^3}} \cdot 15x^2
   \]
   Simplifying, we get:
   \[
   f_x = \frac{15x^2}{2\sqrt{5x^3 + 2y^3}}
   \]

2. **Partial derivative with respect to \( y \):**
   \[
   f_y = \frac{d}{dy} \left( \sqrt{5x^3 + 2y^3} + 2 \right) = \frac{1}{2\sqrt{5x^3 + 2y^3}} \cdot 6y^2
   \]
   Simplifying, we get:
   \[
   f_y = \frac{6y^2}{2\sqrt{5x^3 + 2y^3}} = \frac{3y^2}{\sqrt{5x^3 + 2y^3}}
   \]

### Step 3: Set Partial Derivatives to Zero
For critical points, we set \( f_x = 0 \) and \( f_y = 0 \).

1. **From \( f_x = 0 \):**
   \[
   \frac{15x^2}{2\sqrt{5x^3 + 2y^3}} = 0
   \]
   This implies \( x = 0 \), since the denominator cannot be zero (as it would make the function undefined).

2. **From \( f_y = 0 \):**
   \[
   \frac{3y^2}{\sqrt{5x^3 + 2y^3}} = 0
   \]
   This implies \( y = 0 \), for similar reasons.

### Step 4: Verify the Critical Point
The only solution is \( (x, y) = (0, 0) \). Substituting into the function:
\[
f(0, 0) = \sqrt{5 \cdot 0^3 + 2 \cdot 0^3} + 2 = 2
\]
Thus, \( (0, 0) \) is a critical point.

### Conclusion
The only critical point is:
\[
(x_0, y_0) = (0, 0)
\]

---

Would you like further details on how these derivatives were computed or have any questions? 

#### Related Questions:
1. How do partial derivatives help in finding critical points?
2. Can you explain why setting \( f_x = 0 \) and \( f_y = 0 \) is necessary for critical points?
3. What conditions could lead to no critical points for a function?
4. How would we classify the critical point \( (0, 0) \) in terms of minima, maxima, or saddle point?
5. How do square roots in functions affect differentiability?

#### Tip:
Always check that the function is defined at the critical points you find; certain expressions like square roots restrict the domain.

Find the critical point(s) (x0,y0) of the function f(x,y)=√(5x^3+2y^3) +2. Enter your answer as an ordered pair and separate multiple critical points with commas. If there are no critical points, enter ∅.

Explore how to find the critical points of the multivariable function f(x, y) = √(5x^3 + 2y^3) + 2 by using partial derivatives. Follow the step-by-step solution to identify critical points, setting partial derivatives equal to zero to find potential critical values. Suitable for undergraduate students studying multivariable calculus.

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