Differentiate dTC/dQ = -DS/Q^2 + H/2 with respect to Q again. The first term: d/dQ(-DS * Q^(-2)) = 2 * DS * Q^(-3). The second term H/2 has zero second derivative. Sum: d^2TC/dQ^2 = 2DS/Q^3^[1].

Strictly positive for Q > 0. D and S are positive parameters (demand and order cost are positive by definition). Q is the order quantity, positive by definition. Therefore 2DS/Q^3 > 0 for all feasible Q.

What this proves. The second derivative being strictly positive everywhere means TC(Q) is strictly convex on Q > 0. A strictly convex function has at most one minimum; the stationary point found in step 3 is therefore the global minimum, not a maximum or saddle point^[2].

Why the proof matters. A first-order condition alone gives a stationary point; the second-derivative test is required for rigor. For Wilson EOQ the result is conclusive: Q* is the unique minimum on Q > 0. There is no competing local optimum to worry about, no Q where decreasing further would reduce cost.