Converting Max to Min

Standard form in this course uses minimization.

A maximization problem can be converted into an equivalent minimization problem by multiplying the objective function by $-1$.

Core Rule

The problem:

is equivalent to:

The feasible region is unchanged.

Only the objective sign changes.

Why This Works

If one feasible point has a larger value of $f(x)$, then it has a smaller value of $-f(x)$.

So maximizing $f$ is the same as minimizing $-f$.

If:

then:

Optimal Values

If the original maximum value is:

then the converted minimum value is:

So after solving the minimization problem, convert back by:

Example

Original problem:

subject to some constraints.

Equivalent minimization problem:

subject to the same constraints.

If the minimization problem gives:

then the original maximum value is:

Vector Form

Original maximization:

Equivalent minimization:

The new cost vector is:

Constraints Do Not Change

Converting max to min does not alter:

• $A$
• $b$
• feasible region
• sign restrictions
• equality or inequality direction

Only the objective is negated.

Unboundedness Conversion

If a maximization problem is unbounded above:

then the converted minimization problem is unbounded below:

Common Mistake

Do not multiply the constraints by $-1$ when only converting max to min.

Wrong:

and also changing constraints for no reason.

Correct:

Only the objective changes.

Checklist

You understand this conversion if you can:

• replace $\max f(x)$ by $\min -f(x)$
• keep the feasible region unchanged
• negate the objective coefficients
• convert the optimal value back by changing sign
• distinguish objective conversion from constraint conversion

See Also

• Standard Form
• Objective Functions
• Matrix Form of LP Problems

Exam checkpoint

For standard-form questions, use the course convention $\min c^Tx$ subject to $Ax=b$, $x\ge0$. Convert max objectives, add slack to $\le$, subtract surplus from $\ge$, and split free variables.