Problem regarding a LPP can have a non-basic optimal solution

Question

Which on of the following statements TRUE?

$(A)$ A convex set cannot have infinite many extreme points.

$(B)$ A LPP can have infinite many extreme points.

$(C)$ A LPP can have exactly two different optimal solutions.

$(D)$ A LPP can have a non-basic optimal solution.

Option $(A)$ and $(C)$ are clear. $(A)$ is not true because a closed circle of unit radius is a convex set which have infinite many extreme points. Also $(C)$ is not true because a LPP can have either unique or infinite optimal solutions.

But I am confusing about $(B)$ and $(D)$. I think $(D)$ is not true due to this theorem: If the LPP admits of an optimal solution then the optimal solution will coincide with at least one basic feasible solution of the problem. But answer is $(D)$. If so, where i am wrong. Also provide an example for $(B)$ if $(B)$ is true. Any help is highly appreciated.

Answer 1

A) You got it right.

B) No, as there is a finite number of constraints.

C) You got it right.

D) True: if the problem is unbounded, then one of the variables can take value $\infty$, which does not correspond to a basic solution.

Answer 2

Yes, option b is incorrect because extreme points are corner points of feasible region but option d is incorrect because if optimal solution exists then it must be some finite value and therefore it must b basic feasible solution.

Answer 3

In general, the feasible set of an LPP is a convex polyhedra formed from the intersection of halfspaces (the inequality contraints of the LPP).

If the gradient of the objective is orthogonal to a face or edge of this polyhedra, then a level set of the objective overlaps with the face/edge. This is an example where B is true.