Consider u(x, y) = min{x, y}. Then the preferences represented by u(x, y) are strictly convex.
(a) True (b) False
The statement is false.
A function u(x, y) is strictly convex if, for any two different points (x1, y1) and (x2, y2) in the domain, and for any t between 0 and 1, the following inequality holds:
u(tx1 + (1-t)x2, ty1 + (1-t)y2) < tu(x1, y1) + (1-t)u(x2, y2)
Let's consider the function u(x, y) = min{x, y} and check if it satisfies the definition of strict convexity.
Suppose we have two different points (x1, y1) and (x2, y2) in the domain. Without loss of generality, let's assume that x1 < x2.
Now, let's consider t = 0.5.
tx1 + (1-t)x2 = 0.5x1 + 0.5x2
ty1 + (1-t)y2 = 0.5y1 + 0.5y2
The inequality we need to check is:
u(0.5x1 + 0.5x2, 0.5y1 + 0.5y2) < 0.5u(x1, y1) + 0.5u(x2, y2)
Substituting u(x, y) = min{x, y}, we have:
min{0.5x1 + 0.5x2, 0.5y1 + 0.5y2} < 0.5min{x1, y1} + 0.5min{x2, y2}
Since x1 < x2, we can say that 0.5x1 + 0.5x2 < x1. Similarly, 0.5y1 + 0.5y2 < y1.
Therefore, min{0.5x1 + 0.5x2, 0.5y1 + 0.5y2} = 0.5x1 + 0.5x2
We can also say that 0.5min{x1, y1} + 0.5min{x2, y2} = 0.5x1 + 0.5min{x2, y2}
Since 0.5x1 + 0.5x2 < 0.5x1 + 0.5min{x2, y2}, the inequality is not satisfied.
Therefore, the function u(x, y) = min{x, y} is not strictly convex.
Therefore, the statement is false. The preferences represented by u(x, y) are not strictly convex.