Examine the function for relative extrema. $$ f(x, y)=5 x^{2}+4 y^{2}-10 x-8 y+12 $$

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1. **Find the critical point:**

Compute the first order partial derivatives:
   $$
   f_x(x,y) = \frac{\partial f}{\partial x} = 10x - 10
   $$
   $$
   f_y(x,y) = \frac{\partial f}{\partial y} = 8y - 8
   $$
   
   Set them equal to zero:
   $$
   10x - 10 = 0 \quad \Rightarrow \quad x = 1
   $$
   $$
   8y - 8 = 0 \quad \Rightarrow \quad y = 1
   $$
   
   Therefore, the only critical point is $$(1, 1)$$.

2. **Determine the nature of the critical point using the Hessian matrix:**

Compute the second order partial derivatives:
   $$
   f_{xx} = 10,\quad f_{yy} = 8,\quad f_{xy} = 0
   $$
   
   The Hessian matrix $$ H $$ is:
   $$
   H = \begin{pmatrix}
   10 & 0 \
   0 & 8
   \end{pmatrix}
   $$
   
   Since the eigenvalues of $$ H $$ are $$ 10 $$ and $$ 8 $$, both positive, $$ H $$ is positive definite.

3. **Conclusion:**

Because the Hessian is positive definite at $$(1,1)$$, the function has a relative minimum at this point.
   
   Calculate the function value at this relative minimum:
   $$
   f(1,1) = 5(1)^2 + 4(1)^2 - 10(1) - 8(1) + 12 = 5 + 4 - 10 - 8 + 12 = 3
   $$
   
   Thus, the relative minimum is at $$(1,1)$$ with a value of $$3$$.
The function has a relative minimum at the point (1,1) with a value of 3.

Examine the function for relative extrema. \[ f(x, y)=5 x^{2}+4 y^{2}-10 x-8 y+12 \]

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