f(x,y) = 1 + 4x - 5y, D is the closed triangular region with vertices (0,0), (2,0), and (0,3).
Solution
Find fx\' and fy\' fx\' = 4 (derive f, seeing y as a constant) fy\' = -5 (derive f, seeing
x as a constant) Find an x and y so that: fy\' = 0 and fx\' = 0 But since there is no x or y in the
derivates, it\'s impossible. This means the functions are continously increasing or decreasing.
Therefore the min and max values are as stated in the question: f (2;0) = 9 f (0;3) = -14 Any
other point lies between these values (keep border in mind). So the minimum is -14, the
maximum is 9.
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f(x,y) = 1 + 4x - 5y, D is the closed triangular region with vertice.pdf
1. f(x,y) = 1 + 4x - 5y, D is the closed triangular region with vertices (0,0), (2,0), and (0,3).
Solution
Find fx' and fy' fx' = 4 (derive f, seeing y as a constant) fy' = -5 (derive f, seeing
x as a constant) Find an x and y so that: fy' = 0 and fx' = 0 But since there is no x or y in the
derivates, it's impossible. This means the functions are continously increasing or decreasing.
Therefore the min and max values are as stated in the question: f (2;0) = 9 f (0;3) = -14 Any
other point lies between these values (keep border in mind). So the minimum is -14, the
maximum is 9
