To analyze the maxima and minima of the function

$f(x, y) = x^3 y^2 (1 - x - y)$

we follow these steps:

Step 1: Find the Critical Points

The critical points of $f(x, y)$ occur where the first-order partial derivatives are zero or do not exist.

First-order Partial Derivatives

1. With respect to $x$: $f_x = \frac{\partial}{\partial x} \left( x^3 y^2 (1 - x - y) \right)$ Using the product rule: $f_x = 3x^2 y^2 (1 - x - y) + x^3 y^2 (-1)$ $f_x = 3x^2 y^2 - 3x^3 y^2 - x^3 y^2$ $f_x = 3x^2 y^2 - 4x^3 y^2$ $f_x = x^2 y^2 (3 - 4x)$

2. With respect to $y$: $f_y = \frac{\partial}{\partial y} \left( x^3 y^2 (1 - x - y) \right)$ Again, using the product rule: $f_y = x^3 (2y (1 - x - y) + y^2 (-1))$ $f_y = x^3 (2y - 2xy - 2y^2 - y^2)$ $f_y = x^3 (2y - 2xy - 3y^2)$

Step 2: Solve for Critical Points

Setting both partial derivatives to zero:

1. $f_x = x^2 y^2 (3 - 4x) = 0$

   • This implies $x^2 = 0$, $y^2 = 0$, or $3 - 4x = 0$.
   • So, $x = 0$ or $y = 0$ or $x = \frac{3}{4}$.

2. $f_y = x^3 (2y - 2xy - 3y^2) = 0$

   • This implies $x^3 = 0$ or $2y - 2xy - 3y^2 = 0$.
   • So, $x = 0$ or solving $2y - 2xy - 3y^2 = 0$.

By solving these, the critical points are: $(0,0), (0, \text{any } y), \left(\frac{3}{4}, 0\right), \left(\frac{3}{8}, \frac{1}{4}\right)$

Step 3: Second-Order Partial Derivatives

To determine the nature of these critical points, we compute the Hessian determinant:

$D = f_{xx} f_{yy} - (f_{xy})^2$

where

• $f_{xx} = 6x y^2 - 12x^2 y^2$
• $f_{yy} = x^3 (2 - 6y - 2x)$
• $f_{xy} = 6x^2 y - 12x^3 y - x^3 (2 - 6y - 2x)$

By evaluating $D$ at critical points, we classify them as local minima, maxima, or saddle points.

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Step 4: Conclusion

After analyzing $D$:

• $(0,0)$ is a saddle point.
• $\left(\frac{3}{4}, 0\right)$ is a local maximum.
• $\left(\frac{3}{8}, \frac{1}{4}\right)$ is a local maximum.

Would you like more details on the Hessian determinant calculations?

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Related Questions

1. How do we classify critical points using the second derivative test in multivariable calculus?
2. What is the significance of saddle points in optimization problems?
3. How do we determine global extrema for functions with constrained domains?
4. Can the method of Lagrange multipliers be used to find extrema in this problem?
5. What are real-world applications of maxima and minima in multivariable functions?

Tip:

When solving for critical points, always check the domain of the function, especially when dealing with constraints like $1 - x - y$.