Final answer:
To convert the given absolute value minimization problem into a linear programming problem, break down the absolute value into piecewise linear components and analyze the constraints. The minimum of d(x) = |x - 1| + |x - 3| occurs at x = 1 and x = 3 with a minimum value of 2.
Step-by-step explanation:
The problem given is to minimize d(x) = |x - 1| + |x - 3|, where x is a real number. To convert this problem into a linear programming problem, we need to consider that the objective function d(x) is piecewise linear because of the absolute value functions. The absolute value function |x - a| can be rewritten as two separate conditions: x - a if x ≥ a, and -(x - a) if x < a. Thus, we have two regions to consider, based on the value of x in relation to the points 1 and 3.
The first case is when x is between 1 and 3, i.e., 1 ≤ x ≤ 3. In this region, the function simplifies to d(x) = (x - 1) + (3 - x), because x is greater than 1 and less than 3. The simplified equation is d(x) = 2, which is a constant and therefore its minimum is 2.
In the second case, when x < 1, the function becomes d(x) = (1 - x) + (3 - x), simplifying to d(x) = 4 - 2x. Finally, when x > 3, the function becomes d(x) = (x - 1) + (x - 3), which simplifies to d(x) = 2x - 4.
Now, we consider that the minimum of d(x) will occur at one of the breakpoints, either x = 1 or x = 3, or when the slope of the piecewise linear function changes sign. From the analysis above, we see that the minimum occurs at x = 1 and x = 3 with the minimum value being 2. Therefore, the linear programming problem becomes finding the minimum of a set of linear equations subject to the constraints of our cases: d(x) = 2 when 1 ≤ x ≤ 3, d(x) = 4 - 2x when x < 1, and d(x) = 2x - 4 when x > 3.
When applying graphical methods, we can visually identify the minimum point on a graph representing these three cases. The minimum value found through the graphical method should match our previous analysis that d(x) = 2 at x = 1 and x = 3.