To find the extrema of the function $f(x, y) = x^3 y^2 (1 - x - y)$, we will proceed with the following steps:

1. Compute the First Partial Derivatives

We start by finding the first-order partial derivatives of $f(x, y)$ with respect to $x$ and $y$.

Partial Derivative with respect to $x$:

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

Partial Derivative with respect to $y$:

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

2. Set the Partial Derivatives Equal to Zero

To find the critical points, we set $f_x = 0$ and $f_y = 0$.

Equation for $f_x = 0$:

$x^2 y^2 (3 - 4x - 3y) = 0$ This gives us two cases to consider:

• $x^2 y^2 = 0$ implying $x = 0$ or $y = 0$,
• $3 - 4x - 3y = 0$ simplifying to $4x + 3y = 3$.

Equation for $f_y = 0$:

$x^3 y (2 - 2x - 3y) = 0$ This gives us two cases to consider:

• $x^3 y = 0$ implying $x = 0$ or $y = 0$,
• $2 - 2x - 3y = 0$ simplifying to $2x + 3y = 2$.

3. Solve the System of Equations

We now solve the system of equations formed by:

1. $4x + 3y = 3$
2. $2x + 3y = 2$

Subtract the second equation from the first:

$(4x + 3y) - (2x + 3y) = 3 - 2$ $2x = 1 \quad \Rightarrow \quad x = \frac{1}{2}$

Substitute $x = \frac{1}{2}$ into $2x + 3y = 2$:

$2\left( \frac{1}{2} \right) + 3y = 2$ $1 + 3y = 2 \quad \Rightarrow \quad 3y = 1 \quad \Rightarrow \quad y = \frac{1}{3}$

Thus, the critical point is $(x, y) = \left( \frac{1}{2}, \frac{1}{3} \right)$.

4. Analyze Boundary Conditions

We should also check when either $x = 0$ or $y = 0$, as these are possible solutions due to the conditions in the partial derivatives.

• Case 1: $x = 0$: Substituting into the original function: $f(0, y) = 0$ This gives no critical points unless $y = 0$, where the function also evaluates to 0.

• Case 2: $y = 0$: Substituting into the original function: $f(x, 0) = 0$ Again, this gives no critical points unless $x = 0$, where the function also evaluates to 0.

5. Conclusion

The critical point is $\left( \frac{1}{2}, \frac{1}{3} \right)$. To confirm whether it corresponds to a maximum, minimum, or saddle point, we would need to compute the second-order partial derivatives and apply the second derivative test.

Would you like to continue with the second derivative test, or do you have any questions or clarifications?

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Related Questions:

1. How do you compute the second-order partial derivatives for this function?
2. What is the second derivative test for functions of two variables?
3. How do boundary conditions affect finding extrema in multivariable functions?
4. What are the physical or geometric interpretations of critical points for this function?
5. How can you verify if a critical point is a global extremum?

Tip: When working with multivariable functions, always remember to check both the critical points and boundary conditions.