FlashcardU

Practice: Half-Spaces and Hyperplanes

Recognition · Direct application

A constraint reads 3 x 1 + 2 x 2 ≤ 12 . Where does the point ( 2 , 3 ) lie?

2 hints available, least help first.

Hint 1: Retrieval cue

Substitute the point into the left-hand side and compare the result with 12.

Hint 2: Concept cue

Decide what a closed inequality says about points where the two sides are equal.

Classification · Direct application · Explanation

For each constraint and point below, state whether the point satisfies the constraint, violates it, or lies on its bounding hyperplane. Show the value of the left-hand side in each case, and name the bounding hyperplane.

(a) 4 x 1 + x 2 ≤ 9 at the point ( 2 , 1 )

(b) x 1 − 3 x 2 ≥ 2 at the point ( 1 , 1 )

(c) 2 x 1 + 2 x 2 + x 3 ≤ 10 at the point ( 3 , 1 , 2 )

Write your answer, then compare it with the worked solution.

2 hints available, least help first.

Hint 1: Retrieval cue

For each part, compute the left-hand side at the given point before deciding anything.

Hint 2: Concept cue

The bounding hyperplane is what you get by replacing the inequality sign with an equals sign.

Compare with the worked solution

Comparing does not record a result. Judging your own written answer cannot show that you can do this without help.

(a) 4 ( 2 ) + 1 = 9 , exactly the right-hand side, so ( 2 , 1 ) lies on the bounding hyperplane 4 x 1 + x 2 = 9 . The inequality is closed, so the point satisfies the constraint.

(b) 1 − 3 ( 1 ) = − 2 , and − 2 ≥ 2 is false, so ( 1 , 1 ) violates the constraint. Its bounding hyperplane is x 1 − 3 x 2 = 2 .

(c) 2 ( 3 ) + 2 ( 1 ) + 2 = 10 , equal to the right-hand side, so the point lies on the hyperplane 2 x 1 + 2 x 2 + x 3 = 10 and satisfies the closed inequality. In three variables this hyperplane is a plane, not a line.

The bounding hyperplane. It is the equality case a T x = b . The set where the constraint holds exactly. The inequality divides the space into that boundary and the two sides of it, and naming the boundary is what makes "which side" a well-posed question.

A complete answer does each of these:

  • evaluates constraint
  • correct side
  • boundary included
  • names hyperplane

Error diagnosis · Explanation · Transfer

A student writes:

Each constraint of a linear program is a line, so with five constraints in four variables I can draw the five lines and read off the feasible region. A point exactly on a line is not in the region, because the region is what the lines enclose.

There are two separate errors here. Identify both, correct each one, and state what the constraint x 1 + x 2 + x 3 + x 4 ≤ 7 actually describes in R 4 .

Write your answer, then compare it with the worked solution.

2 hints available, least help first.

Hint 1: Retrieval cue

How many dimensions does one linear equation remove from the space it is written in?

Hint 2: Concept cue

Consider separately what the student says about shape and what they say about points lying exactly on a boundary.

Compare with the worked solution

Comparing does not record a result. Judging your own written answer cannot show that you can do this without help.

First error: the dimension. A single linear equation in n variables describes a flat of dimension n − 1 , not a line. A line is what that happens to be when n = 2 . In four variables the equation x 1 + x 2 + x 3 + x 4 = 7 describes a three-dimensional hyperplane, and the constraint x 1 + x 2 + x 3 + x 4 ≤ 7 describes the closed half-space on one side of it. Neither can be drawn, and the plan to read the region off a sketch does not survive past two variables.

Second error: the boundary. A closed inequality includes the points where the two sides are equal, so a point on the bounding hyperplane is in the feasible region rather than excluded from it. This matters beyond bookkeeping: optima in linear programming characteristically lie on constraint boundaries, so a rule that discards them discards the answers.

What survives the move to higher dimensions is the arithmetic. To test a point, evaluate the left-hand side and compare it with the right; the procedure is identical in two variables and in four hundred.

Naming the boundary. The bounding hyperplane of a T x ≤ b is a T x = b . A point there satisfies a closed inequality; it is on the boundary of the half-space, not outside it, which is the distinction the error above turns on.

A complete answer does each of these:

  • evaluates constraint
  • correct side
  • boundary included
  • names hyperplane
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