“Minimise the cost” needs calculus. “Is there an order quantity costing at most T” needs a quadratic: multiply K·D/Q + h·Q/2 ≤ T through by Q and it becomes h·Q²/2 − T·Q + K·D ≤ 0, which holds exactly between the two roots. Lower T and the roots move together; they meet when the discriminant reaches exactly zero, and the single quantity left at that moment is the economic order quantity. Existence, uniqueness and optimality arrive together, and nothing is differentiated.
Core ideas
Three ideas. The first is the change of question that makes the algebra possible, the second is the step that has a condition attached, and the third is what the collision actually proves.
“What is the smallest cost” is a search over all quantities. “Is there a quantity costing at most T” is a yes-or-no question about one inequality, and inequalities in one variable with a squared term in them are something Algebra already solves. Lowering T until the answer turns from yes to no is then a complete method, and the turning point is exact rather than approached.
Multiplying both sides by a negative number reverses an inequality. The step from K·D/Q + h·Q/2 ≤ T to h·Q²/2 − T·Q + K·D ≤ 0 multiplies by Q, and it keeps the direction only because an order quantity is positive. That is not pedantry: it is the one place in this derivation where a condition on the variable is doing real work, and the lab refuses a zero or negative Q for the same reason.
While the discriminant is positive there are many quantities cheap enough, and the budget is not tight. When it is negative there are none, and the budget is impossible. Exactly one budget separates those two cases, the interval of affordable quantities has shrunk to a single point there, and that point is therefore the unique cheapest quantity. Nothing else can be: any cheaper one would make the discriminant negative and it would not exist.
The material
The rearrangement, the three regimes of the discriminant, and the collision that produces the answer.
Fix the model and ask a different question. Instead of “what is the cheapest order quantity”, ask “is there an order quantity costing at most T per unit time”. That is K·D/Q + h·Q/2 ≤ T with Q > 0, and multiplying through by the positive quantity Q turns it into a polynomial inequality:
For the quadratic h·Q²/2 − T·Q + K·D the discriminant is T² − 4·(h/2)·(K·D) = T² − 2hKD. Its sign decides everything:
positive, and there is an interval of order quantities meeting the budget; zero, and there is exactly one; negative, and there is none — the parabola never reaches the axis and no quantity is that cheap.
So the budget can be walked down. Each step is a fresh quadratic with a fresh discriminant, and the interval of affordable quantities closes as the budget tightens. The panel does exactly that, on K = 100, D = 1200, h = 6, where the quadratic is 3Q² − T·Q + 120000 and 2hKD = 1440000.
T² − 2hKD = 0 exactly when T = √(2hKD). At that budget the two roots coincide at Q = T/h = √(2hKD)/h = √(2KD/h).
No smaller budget is met by any quantity, since the discriminant is then negative; and that single Q meets this one. So √(2hKD) is the least achievable cost and √(2KD/h) is the unique quantity achieving it.
That is the whole derivation. It uses the quadratic formula, the sign of a leading coefficient, and the fact that a positive variable may be multiplied through an inequality. It proves existence, uniqueness and optimality in one motion, and it never mentions a rate of change. Where a calculus treatment differentiates C(Q), sets the result to zero and then checks a second derivative to rule out a maximum, this argument has no second case to rule out: an upward parabola cannot be below zero outside its roots.
Set T = 1300 on the worked instance. The discriminant is 1300² − 1440000 = 250000, a perfect square, and the roots are 400/3 and 300 exactly. Every order quantity between them costs at most 1300: the panel prices 134, 200 and 299 and all three clear the budget, while 133 and 301 do not.
Both endpoints cost exactly 1300 — they are where the cost curve crosses the budget line — so the interval is closed at both ends, which is what ≤ in the original question asked for.
T = 1199 gives 1199² − 1440000 = −2399. The parabola 3Q² − 1199Q + 120000 stays strictly above the axis, and the panel draws it in red with no shaded interval at all.
This is the informative failure. The answer is not “the arithmetic went wrong” but “no order quantity whatsoever costs as little as 1199 per unit time on this data”, and the discriminant is the proof of it.
Setting the budget as a percentage of C* connects this page to the flat band of “The Economic Order Quantity”, because they are the same interval computed two ways. At T = 1260, which is 105% of 1200, the roots are 210 ± 10√41 ≈ 145.9688 … 274.0312; the flatness band at a 5% tolerance is 145.9688 … 274.0312. At 110% the roots are 220 ± 20√21 ≈ 128.3485 … 311.6515 and the 10% band is 128.3485 … 311.6515. One is a discriminant in Q, the other a discriminant in t, and they agree because they are the same question.
The panel's budget slider is a percentage of C*, and where C* is irrational that percentage is applied to a rounded decimal of it. The argument itself never touches that decimal — it runs on whatever rational T the slider produced — but it does mean that on an irrational instance the 100% row is very close to a merge rather than exactly one, and the panel says so rather than pretending the roots met. On the worked instance C* = 1200 exactly, so the merge really is a merge and the discriminant prints as exactly zero.
One consequence is worth carrying into the rest of the course. The width of the affordable interval shrinks to zero, but it shrinks slowly: at a budget 5% above the minimum the interval is still 128 units wide out of an optimum of 200. That is the flatness of the cost curve seen from underneath, and it is why the models in the rest of this course can afford to round their answers to a case size, a pallet or a whole number of units.
Interactive
no derivative: a quadratic, its discriminant, and the budget at which the interval closes
Method
Five steps. The third is where most of the errors live, and it is the one that costs nothing to check.
K·D/Q + h·Q/2 ≤ T. Not an equation: the question is whether anything is cheap enough, and the set of answers is generally an interval rather than a point.
Q > 0, so the inequality keeps its direction and becomes h·Q²/2 − T·Q + K·D ≤ 0. Write the reason down. A reader who omits it has the right answer by accident and will lose it the first time a variable can be negative.
h/2 > 0, so the parabola opens upward and the inequality holds BETWEEN the roots. If it opened downward the solution set would be everything outside them — two rays, no bounded interval, and no collision to find.
Positive: an interval of affordable quantities, ends at (T ± √(T² − 2hKD))/h. Zero: exactly one, at T/h. Negative: none, and the budget is impossible rather than merely demanding.
T = √(2hKD) is the least achievable cost and T/h = √(2KD/h) is the quantity that achieves it. Keep both as surds; the decimal is a printing step, and doing it earlier makes the merge look approximate when it is not.
K = 100, D = 1200, h = 6, so the quadratic is 3Q² − T·Q + 120000 and 2hKD = 1440000. Every root below is what the panel reports, and the surds are the panel's own.
The last line is the only place the cost function appears, and it is a check rather than a step. Everything above it came from factorising a quadratic, which is why the argument survives the path's promise of no calculus: 1200 and 200 were produced by the discriminant and then confirmed by the thing they are supposed to minimise.
Notice that T = 1300 gives rational roots and T = 1260 does not, on the same instance. Whether the ends of the interval are nice has nothing to do with whether the optimum is: 1300² − 1440000 = 250000 = 500² is luck, while 1200² − 1440000 = 0 is the theorem. A page that printed both as decimals would have hidden the difference.
For a rehearsal, take K = 50, D = 600, h = 5, where 2hKD = 300000 and C* = 100√30. The first move is given: the merge budget is irrational, so the panel's percentage slider is working from a rounded 547.7226 and the 100% row shows a very small positive discriminant instead of zero. Say what the two roots look like just before they meet, and why the panel is right to report that rather than to round the discriminant to zero.
You should be able to turn a cost target into a quadratic inequality with the positivity condition stated, read all three regimes off the discriminant, locate the budget at which the roots merge, derive Q* = √(2KD/h) and C* = √(2KDh) from that merge, and explain why the collision proves optimality without any appeal to a rate of change.
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