Contour Lines and the Direction of Improvement

A maximisation with positive coefficients. $max3x_1+2x_2$ gives $c=(3,2)$. Contours have slope $-3/2$. Push along $(3,2)$, up and to the right. Check: from $(0,0)$ to $(3,2)$ the objective goes from $0$ to $13$.

The same objective minimised. $min3x_1+2x_2$. The contours are identical lines with identical slope. Push against $(3,2)$, towards the origin. Nothing about the drawing changed; the travel direction reversed.

A negative coefficient. $max2x_1-5x_2$ gives $c=(2,-5)$. Improvement means moving right and down. A learner pushing 'away from the origin' by habit would slide up and to the right and lose $5$ per unit of $x_2$ gained.

A direction test. With $c=(2,-5)$, is $d=(1,1)$ improving for a maximisation? : no, it worsens the objective. Is $d=(5,2)$? $c^{\mathsf{T}}d=10-10=0$: neither. It runs along a contour, and the objective is unchanged.

The last case is the significant one. A direction with $c^{\mathsf{T}}d=0$ is exactly how a linear program comes to have an edge of equally good answers.

Not the improving direction: the contour's own slope. The contour for $max3x_1+2x_2$ has slope $-3/2$, and $(-2,3)$ is a direction along it. Moving that way changes the objective by $c^{\mathsf{T}}d=-6+6=0$. Along the contour is precisely where nothing improves.

Not the improving direction: away from the origin. For $max2x_1-5x_2$, the direction $(1,1)$ leads away from the origin and worsens the objective, since $c^{\mathsf{T}}(1,1)=-3$. The habit works only for objectives with all-positive coefficients and fails quietly otherwise.

Not a contour: a constraint line. $x_1+2x_2\le 8$ drawn as $x_1+2x_2=8$ is a boundary of the region. It bounds what is permitted. A contour bounds nothing. It records where the objective is constant, and it extends across the whole plane, inside the region and outside it alike.

Not a reason to reverse $c$: a minimisation. For $min3x_1+2x_2$ the vector $c=(3,2)$ still points towards increase. It is not redefined as $(-3,-2)$. What changes is that you travel against it. Redefining $c$ per problem type is how sign errors enter the reduced-cost test later, where there is no picture to catch them.

Two facts about a linear objective look like one fact, and separating them is the whole of this unit.

What the drawing settles. Where the equal-value lines run. This needs only the ratio of the coefficients, and it is identical for $max3x_1+2x_2$, for $min3x_1+2x_2$, and for $max-3x_1-2x_2$. All three produce the same family of parallel lines of slope $-3/2$.

What the drawing cannot settle. Which way along them improves. Those three problems travel in three different ways: along $(3,2)$, against $(3,2)$, and against $(3,2)$ respectively. A family of parallel lines carries no orientation, so no amount of care in drawing recovers it.

Why the error survives inspection. Suppose the direction is taken backwards on the first of these. The contours are correct, the region is correct, the contour is slid neatly until it last touches, at the wrong corner. The output is a specific point and a specific value, both plausible, neither flagged by anything visible. Compare an arithmetic slip, which usually produces something that looks odd.

The habit that prevents it. Write $c$ down explicitly, state whether you are maximising or minimising, and say the travel direction in words before sliding anything: maximising, so travel along $c$, or minimising, so travel against $c$. When a direction is in doubt, test it: compute $c^{\mathsf{T}}d$ and read the sign. One multiplication settles what an hour of careful drawing cannot.

The direction in which to push the objective contour can be read from the drawn contour line itself, so a correctly drawn contour family determines which way improves.