Let Y = minimum of {(x+2), (3-x)}. What is the maximum value of Y for 0 <= x <=1?

Solution

For x<0 ; y=x+2

for 0<x<$$\frac{1}{2}$$ ; $$y=x+2$$

for $$x>\frac{1}{2}$$ ; $$y=3-x$$

Hence, $$y$$ attains its maxima at $$x=\frac{1}{2}$$ i.e. $$y$$ = 2.5

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Let Y = minimum of {(x+2), (3-x)}. What is the maximum value of Y for 0 <= x <=1?

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