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: $(x_0, y_0) = (0, 0)$

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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.