Question

Let $Y = \min((x + 2), (3 - x))$. What is the maximum value of $Y$ for $0 \le x \le 1$?

• A. 1.0
• B. 1.5
• C. 3.1
• D. 2.5

Hint:

When optimizing a piecewise function with min or max, find crossover point and test all boundary values.

Answer 1

The Correct Option is B

Step 1: Understand behavior of each function \[ f_1(x) = x + 2,\quad f_2(x) = 3 - x \] We want: \[ Y = \min(x+2,\ 3-x) \] Step 2: Equate to find intersection point \[ x + 2 = 3 - x \Rightarrow 2x = 1 \Rightarrow x = 0.5 \] So for $x<0.5$: $x + 2<3 - x \Rightarrow Y = x + 2$ For $x>0.5$: $3 - x<x + 2 \Rightarrow Y = 3 - x$ Step 3: Find max Y in interval [0, 1] Max Y occurs at $x = 0.5$ Then: $Y = x + 2 = 0.5 + 2 = 2.5$ and $3 - x = 2.5$ So minimum is 2.5 \[ \boxed{1.5} \quad \text{(Wait! That contradicts)} \] Actually, at $x=0.5$: \[ Y = \min(2.5, 2.5) = 2.5 \Rightarrow \boxed{2.5} \] Correct option is: \[ \boxed{2.5} \]