State the conditions of unconstrained optimization and find maxima and minima of the function, U = 36 - 6x ^ 2 - 3y ^ 2 + 12x + 10y.solution

### Conditions of Unconstrained Optimization

For an unconstrained optimization problem, we seek to find the maxima or minima of a function without any external constraints on the variables. The necessary conditions for optimization are:

1. **First-Order Condition (FOC)**: The first derivatives (partial derivatives for multivariable functions) of the function must equal zero. This gives the **critical points** of the function.
   
   \[
   \frac{\partial f}{\partial x} = 0 \quad \text{and} \quad \frac{\partial f}{\partial y} = 0
   \]

2. **Second-Order Condition (SOC)**: The second derivatives (second partial derivatives) are used to determine the nature of the critical points (whether they are maxima, minima, or saddle points). This is done using the **Hessian matrix**:
   
   \[
   H = \begin{pmatrix}
   \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \\
   \frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2}
   \end{pmatrix}
   \]
   The nature of the critical points depends on the determinant of the Hessian matrix, \( \det(H) \), and the signs of the second derivatives:
   - If \( \det(H) > 0 \) and \( \frac{\partial^2 f}{\partial x^2} > 0 \), there is a local minimum.
   - If \( \det(H) > 0 \) and \( \frac{\partial^2 f}{\partial x^2} < 0 \), there is a local maximum.
   - If \( \det(H) < 0 \), the point is a saddle point.

### Step-by-Step Solution for the Function \( U(x, y) = 36 - 6x^2 - 3y^2 + 12x + 10y \)

1. **First-Order Conditions (FOC)**:
   \[
   U(x, y) = 36 - 6x^2 - 3y^2 + 12x + 10y
   \]
   Take the partial derivatives with respect to \(x\) and \(y\):

   \[
   \frac{\partial U}{\partial x} = -12x + 12 = 0
   \]
   \[
   \frac{\partial U}{\partial y} = -6y + 10 = 0
   \]

   Solving these equations:
   \[
   -12x + 12 = 0 \quad \Rightarrow \quad x = 1
   \]
   \[
   -6y + 10 = 0 \quad \Rightarrow \quad y = \frac{5}{3}
   \]

   So, the critical point is \( (x, y) = \left( 1, \frac{5}{3} \right) \).

2. **Second-Order Conditions (SOC)**:
   Compute the second partial derivatives:

   \[
   \frac{\partial^2 U}{\partial x^2} = -12, \quad \frac{\partial^2 U}{\partial y^2} = -6, \quad \frac{\partial^2 U}{\partial x \partial y} = 0
   \]

   The Hessian matrix is:
   \[
   H = \begin{pmatrix}
   -12 & 0 \\
   0 & -6
   \end{pmatrix}
   \]
   The determinant of the Hessian is:
   \[
   \det(H) = (-12)(-6) - (0)(0) = 72
   \]

   Since \( \det(H) > 0 \) and \( \frac{\partial^2 U}{\partial x^2} = -12 < 0 \), the critical point \( \left( 1, \frac{5}{3} \right) \) is a **local maximum**.

### Conclusion:

- The function \( U(x, y) = 36 - 6x^2 - 3y^2 + 12x + 10y \) has a **local maximum** at \( \left( x, y \right) = \left( 1, \frac{5}{3} \right) \).

Do you have any questions, or would you like further details?

### Related Questions:
1. What is the difference between constrained and unconstrained optimization?
2. How can you interpret critical points geometrically in optimization?
3. Can we apply this method to higher-dimensional functions?
4. What happens if the determinant of the Hessian is zero?
5. How does the optimization process change for non-differentiable functions?

### Tip:  
Always check the second-order condition to confirm whether a critical point is a maximum, minimum, or saddle point.

State the conditions of unconstrained optimization and find maxima and minima of the function, U = 36 - 6x^2 - 3y^2 + 12x + 10y.

Explore the conditions of unconstrained optimization to find the maxima and minima of the multivariable function U = 36 - 6x^2 - 3y^2 + 12x + 10y. Learn about first-order and second-order conditions, including the use of the Hessian matrix, and determine the nature of critical points. Suitable for undergraduate mathematics students.

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