If $x \geq 2, y - x \geq - 3, and x + y \leq 5$ $x>=2, y-x>=-3, and x+y<=5$ , what is the maximum value of $f (x, y) = x - 4 y$ $f(x,y)=x-4y$ ?

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Answer 1 · 2018-03-12T09:25:43

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First, we have to draw the interception of three given conditions. We have something like this:

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Where the intercpetions points are $(4, 1); (2, 3) and (2, - 1)$ $(4,1);(2,3) and (2,-1)$

By a famous theorem, we know that maximum values of a linear funtion lies in intersection points of restriction area. Thus we proof with this values

For $(4, 1)$ $(4,1)$ ; $f (x, y) = x - 4 y = 4 - 4 \cdot 1 = 0$ $f(x,y)=x-4y=4-4·1=0$

For $(2, 3)$ $(2,3)$ ; $f (x, y) = x - 4 y = 2 - 4 \cdot 3 = - 10$ $f(x,y)=x-4y=2-4·3=-10$

For $(2, - 1)$ $(2,-1)$ ; $f (x, y) = x - 4 y = 2 + 4 \cdot 1 = 6$ $f(x,y)=x-4y=2+4·1=6$

So, the maximum value occurs in $(2, - 1)$ $(2,-1)$

If x≥2,y−x≥−3,andx+y≤5x>=2, y-x>=-3, and x+y<=5, what is the maximum value of f(x,y)=x−4yf(x,y)=x-4y?