Minimization-to-Maximization Transformation¶

Proposition¶

Over the same feasible set, minimizing $f(\boldsymbol{x})$ is equivalent to maximizing $-f(\boldsymbol{x})$. The two problems have the same optimal solution vectors, and their optimal objective values have opposite signs.

Interpretation¶

Minimizing and maximizing are mirror images: a feasible point where $f$ is smallest is exactly a point where $-f$ is largest. The simplex method is presented for maximization, and negating a minimization objective lets the same procedure be used while the constraints stay unchanged.

The red curve $-f$ is the blue curve $f$ reflected in the horizontal axis. The lowest point of $f$ and the highest point of $-f$ lie above and below the same $x^\ast$, at heights $f(x^\ast)$ and $-f(x^\ast)$.

The reason is a single fact about inequalities: multiplying by $-1$ reverses the direction, so $f(\boldsymbol{x}^\ast)\le f(\boldsymbol{x})$ holds exactly when $-f(\boldsymbol{x}^\ast)\ge-f(\boldsymbol{x})$$-f(\mathbf{x})$. Only the objective is negated; the feasible set stays the same, and the optimal values satisfy $\min f=-\max(-f)$$-\max (-f)$.

Two slips are common: negating the constraints as well, which changes the feasible set, and forgetting to change the sign back at the end, so that the transformed value 10 is reported instead of the original minimum $-10$.

Key Results¶

• Negate every objective coefficient.
• Recover the original optimal value by changing the sign back.
• Only the objective changes; the allowed decisions remain the same.

Examples¶

• Minimizing $3x_1-5x_2$ becomes maximizing $-3x_1+5x_2$$5x_2$.
• In week 2 practice 8, transformed value 10 corresponds to original minimum $-10$.

Prerequisites¶

• Objective Function
• Optimal Solution

Related Concepts¶

• Simplex Method
• Assignment Cost Transformation

Sources¶

• OR Modelling · Week 2

Last updated: 20 October 2026 03:02