Question

z=f(x,y)=(x−4) 2 +(y−2) 2 +40 subject to 0≤x≤10 and 0≤y≤15 a) Sketch the bounded region with the constraints above and follow the steps below to complete the problem b) Find all the corner points (with their coordinates) for the region. c) Find any critical point(s) (with their coordinates) that lie within the region. d) Find all the critical points (with their coordinates) when subjecting f(x,y) to the boundary constraints. e) Find all corresponding z or functional values for each of the coordinate points from (b), (c), and (d) in a chart and state the Absolute Maximum and Absolute Minimum values of the function and their corresponding (x,y) points.

Answer 1

The absolute minimum value of the function
`\( f(x, y) = (x - 4)^2 + (y - 2)^2 + 40 \)` within the given constraints occurs at the point (4, 2) with a minimum value of 40. The absolute maximum value is obtained at the corner point (10, 15) with a maximum value of 310.

Step-by-step explanation:

To find the corner points, we evaluate the function at the vertices of the feasible region determined by the constraints
`\(0 \leq x \leq 10\) and \(0 \leq y \leq 15\). The corner points are (0, 0), (0, 15), (10, 0), and (10, 15). Next, we check for critical points within the region by finding the partial derivatives of \( f(x, y) \) and solving for \( (\partial f)/(\partial x) = 0 \) and \( (\partial f)/(\partial y) = 0 \).`However, in this case, there are no critical points within the interior of the region.

Moving on to the boundary constraints, we evaluate
`\( f(x, y) \) along the lines \( x = 0 \), \( x = 10 \), \( y = 0 \), and \( y = 15 \)`, finding critical points at (0, 2), (10, 2), (4, 0), and (4, 15). Finally, we compare the functional values at all the obtained points, concluding that the absolute minimum occurs at (4, 2) with a value of 40, and the absolute maximum occurs at (10, 15) with a value of 310.