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

To find the relative extrema of the function \( f(x, y) = 5x^2 + 4y^2 - 10x - 8y + 12 \), you first need to compute the partial derivatives \( f_x \) and \( f_y \). Setting both derivatives to zero will help locate critical points. The calculations yield: \[ f_x = 10x - 10 = 0 \quad \Rightarrow \quad x = 1 \] \[ f_y = 8y - 8 = 0 \quad \Rightarrow \quad y = 1 \] So the critical point is at \( (1, 1) \). Next, to determine the nature of this critical point, compute the second partial derivatives and utilize the second derivative test. To proceed with the fun part: At the critical point \( (1, 1) \), calculate the Hessian determinant \( D = f_{xx} f_{yy} - (f_{xy})^2 \), where \( f_{xx} = 10 \), \( f_{yy} = 8 \), and \( f_{xy} = 0 \). Thus, \( D = (10)(8) - (0)^2 = 80 \). Since \( D > 0 \) and \( f_{xx} > 0 \), the point \( (1, 1) \) is a local minimum! Grab a pencil and sketch that paraboloid to see how it beautifully dips down at this point! Now for a bit of practical application: functions like this one are commonly used in optimization problems across various fields such as economics, engineering, and physics. Whether you're looking to minimize costs or maximize efficiency, understanding how to find these critical points can lead to valuable insights and solutions. So the next time you’re planning a budget or designing an experiment, remember, a few equations can lead to a treasure trove of optimization!