Corrected exercises: applied optimization (201-SN2-RE)

Diagram: a rectangle with one side of length $L$ along the barn wall, and the other three sides fenced, namely the side $L$ opposite the wall and the two sides of width $\ell$. Function to OPTIMIZE: the area $A=L\ell$. CONSTRAINT: the available fencing gives $L+2\ell=200$. Translating the statement decides everything: mixing up the two relations, or fencing four sides when the wall spares one, brings down the whole rest of the problem even if the computation that follows is flawless.

From the constraint, $L=200-2\ell$, which we substitute into the area: $A(\ell)=(200-2\ell)\ell=200\ell-2\ell^{2}$. Domain: we need $\ell>0$ and $L=200-2\ell>0$, that is, $\ell<100$. The admissible domain is therefore $\left(0, 100\right)$, or $\left[0, 100\right]$ if we accept the degenerate cases of zero area. Taking the CLOSED interval $\left[0, 100\right]$ is convenient here, because it lets us invoke the Extreme Value Theorem directly in part c).

$A'(\ell)=200-4\ell$, which vanishes for $\ell=50$. Then $L=200-100=100$ and the area is $A=100\times 50=5000$ m². Justification: on the closed interval $\left[0, 100\right]$, the continuous function $A$ reaches its absolute maximum, and the candidates are the critical number $\ell=50$ and the endpoints. Now $A(0)=0$ and $A(100)=0$, while $A(50)=5000$: the absolute maximum is indeed reached at $\ell=50$ ✓. We could also conclude from $A''(\ell)=-4<0$, which shows a parabola that is concave down, so a vertex that is a maximum. The optimal dimensions are $100$ m parallel to the wall and $50$ m perpendicular to it.

With four fenced sides, the constraint becomes $2L+2\ell=200$, that is, $L+\ell=100$ and $L=100-\ell$. The area is $A(\ell)=(100-\ell)\ell$, whose derivative $100-2\ell$ vanishes at $\ell=50$, hence $L=50$: we get a SQUARE with sides of 50 m and an area of $2500$ m². Comparison: with the same length of fencing, the pen against the wall reaches $5000$ m², exactly TWICE as much. The optimal shape changes too: a square in the second case, a rectangle twice as long as it is wide in the first. The explanation fits in one sentence: the wall provides one side for free, and the optimum always consists in sharing the fencing evenly between the two DIRECTIONS, here 100 m for the direction parallel to the wall and $2\times 50$ m for the perpendicular direction. This principle of balance between directions comes back in almost every geometric optimization problem.

Let $x$ be the side of the square base and $h$ the height. Volume (CONSTRAINT): $V=x^{2}h=32{,}000$, hence $h=\frac{32{,}000}{x^{2}}$. Surface area to MINIMIZE, with no lid, so one base and four walls: $S=x^{2}+4xh$. Substituting: $S(x)=x^{2}+4x\cdot\frac{32{,}000}{x^{2}}=x^{2}+\frac{128{,}000}{x}$.

Domain: $x>0$ strictly, since $x=0$ defines no box and would make $h$ infinite. On the other hand there is no upper bound: a huge base with a tiny height is still geometrically admissible. The domain is therefore $\left(0, +\infty\right)$, an OPEN and unbounded interval. The Extreme Value Theorem does not apply, since it requires a closed and bounded interval: here it guarantees the existence of no extremum at all, and the justification will have to come from elsewhere. Note also that $S\to+\infty$ at both ends: when $x\to 0^{+}$ because of the term $\frac{128{,}000}{x}$, and when $x\to+\infty$ because of $x^{2}$.

$S'(x)=2x-\frac{128{,}000}{x^{2}}$. Setting it to zero: $2x=\frac{128{,}000}{x^{2}}$, so $2x^{3}=128{,}000$, that is, $x^{3}=64{,}000$ and $x=40$ cm. Then $h=\frac{32{,}000}{1600}=20$ cm, and the surface area is $S=1600+\frac{128{,}000}{40}=1600+3200=4800$ cm². Justification valid on an open interval: $S''(x)=2+\frac{256{,}000}{x^{3}}>0$ for every $x>0$, so $S$ is CONCAVE UP on the whole domain; a critical point of a concave up function is necessarily a minimum, and it is absolute. We could also argue with the First Derivative Test: $S'<0$ for $x<40$ and $S'>0$ for $x>40$, so $S$ decreases then increases, with a single critical number: the minimum at $x=40$ is therefore absolute. This is the expected argument, and it replaces any appeal to the endpoints.

Without a lid we get $h=20$ and $x=40$, that is, $\frac{h}{x}=\frac{1}{2}$: the box is twice as wide as it is tall. With a lid, the surface area would be $S=2x^{2}+4xh=2x^{2}+\frac{128{,}000}{x}$, with derivative $4x-\frac{128{,}000}{x^{2}}$, which vanishes for $x^{3}=32{,}000$, that is, $x\approx 31.75$ cm and $h=\frac{32{,}000}{x^{2}}\approx 31.75$ cm: we get a CUBE, with $\frac{h}{x}=1$. The result is intuitive in hindsight: with a lid, all directions play the same role and the most economical shape is the cube, the three-dimensional analogue of the square in the previous problem. Without a lid, the top is free, so it pays to SPREAD the box out to enlarge the face that costs nothing, exactly as the barn wall made it pay to stretch the pen along its direction. The same principle governs both problems: the dimension whose surface is not charged for should be developed.

Since the cylinder is inscribed and centred, its axis passes through the centre of the sphere and its two base circles lie at a distance $\frac{h}{2}$ from the centre. A radius of the sphere joining the centre to a point on the rim of a base forms a right triangle with legs $r$ and $\frac{h}{2}$ and hypotenuse $R$. The constraint is therefore $r^{2}+\left(\frac{h}{2}\right)^{2}=R^{2}$, here $r^{2}+\frac{h^{2}}{4}=9$.

Volume of the cylinder: $V=\pi r^{2}h$. Since the constraint gives $r^{2}=9-\frac{h^{2}}{4}$ directly, the substitution is immediate and avoids any square root: $V(h)=\pi\left(9-\frac{h^{2}}{4}\right)h=\pi\left(9h-\frac{h^{3}}{4}\right)$. Domain: we need $h>0$ and $r^{2}>0$, that is, $\frac{h^{2}}{4}<9$, or $h<6$. The domain is therefore $\left(0, 6\right)$, which can be closed to $\left[0, 6\right]$ to include the degenerate cases of zero volume. Substituting $r^{2}$ rather than $r$ is the technical move of the exercise: writing $r=\sqrt{9-\frac{h^{2}}{4}}$ and substituting it into $V$ would lead to the same result, but with a much more painful differentiation.

$V'(h)=\pi\left(9-\frac{3h^{2}}{4}\right)$, which vanishes when $\frac{3h^{2}}{4}=9$, that is, $h^{2}=12$ and $h=2\sqrt{3}\approx 3.464$. Then $r^{2}=9-\frac{12}{4}=6$, so $r=\sqrt{6}\approx 2.449$. Volume: $V=\pi\cdot 6\cdot 2\sqrt{3}=12\sqrt{3}\,\pi\approx 65.30$. Justification: on $\left[0, 6\right]$, $V(0)=V(6)=0$ and $V\left(2\sqrt{3}\right)>0$, so the absolute maximum is indeed at the critical number ✓ (we can also note that $V''(h)=-\frac{3\pi h}{2}<0$ on the domain, hence concave down and a maximum).

Volume of the sphere: $V_{S}=\frac{4}{3}\pi R^{3}=\frac{4}{3}\pi(27)=36\pi\approx 113.10$. Ratio: $\frac{12\sqrt{3}\pi}{36\pi}=\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}\approx 0.577$, that is, about $57.7$ percent. The ratio does NOT depend on $R$, and you can see it without redoing the computation: the problem is invariant under a change of scale, since multiplying all lengths by a factor $k$ multiplies all volumes by $k^{3}$, which leaves their ratio unchanged. Redoing the computation with any $R$, we would indeed find $h=\frac{2R}{\sqrt{3}}$ and $r^{2}=\frac{2R^{2}}{3}$, hence $V=\frac{4\pi R^{3}}{3\sqrt{3}}$ and the ratio $\frac{1}{\sqrt{3}}$ ✓. This homogeneity argument pays off on an exam: it predicts that a dimensionless ratio will be universal, and checks the consistency of the result along the way.

After cutting and folding, the height of the box is $x$ (the folded-up square), and the base is a square whose side has lost $x$ at each end, so it measures $30-2x$. The volume is $V(x)=x(30-2x)^{2}$.

We need $x>0$ for the cut to exist, and $30-2x>0$ for the base to survive, that is, $x<15$. The admissible domain is therefore $\left(0, 15\right)$, which can be closed to $\left[0, 15\right]$ by accepting the two degenerate cases of zero volume (no cut at all, or a cut that uses up the whole sheet). Closing the interval is convenient here, because it lets us invoke the Extreme Value Theorem.

Let us expand before differentiating, it is safer: $V(x)=x\left(900-120x+4x^{2}\right)=900x-120x^{2}+4x^{3}$, so $V'(x)=900-240x+12x^{2}=12\left(x^{2}-20x+75\right)=12(x-5)(x-15)$. The critical numbers are $x=5$ and $x=15$. On $\left[0, 15\right]$, the candidates are therefore $0$, $5$ and $15$: $V(0)=0$, $V(15)=0$ and $V(5)=5(30-10)^{2}=5(400)=2000$ cm³. The absolute maximum is therefore $2000$ cm³, reached for a cut of $5$ cm ✓. You can also factor $V'(x)=(30-2x)(30-6x)$ directly, which gives the roots without expanding.

The second critical number $x=15$ corresponds to the endpoint of the domain: cutting 15 cm from each corner of a 30 cm sheet uses up the whole width, the base disappears and the volume is zero. It is therefore not a maximum but the degenerate case, mathematically valid (it does make the derivative zero) and physically meaningless. It is a useful reminder: solving $V'(x)=0$ produces CANDIDATES, and you must always confront them with the domain and with geometric common sense before concluding. A student who announced $x=15$ without evaluating $V$ would end up with a box of zero volume.

The width of the rectangle is $2r$ (the diameter of the semicircle). The perimeter consists of the base $2r$, the two vertical sides $2h$, and the arc of the semicircle $\pi r$ (half of $2\pi r$): the upper horizontal side does not exist, since the semicircle sits on it. CONSTRAINT: $2r+2h+\pi r=8$. AREA: that of the rectangle plus that of the semicircle, $\mathcal{A}=2rh+\frac{\pi r^{2}}{2}$.

From the constraint, $2h=8-2r-\pi r$, so $h=\frac{8-2r-\pi r}{2}$. Substituting: $\mathcal{A}(r)=2r\cdot\frac{8-2r-\pi r}{2}+\frac{\pi r^{2}}{2}=r(8-2r-\pi r)+\frac{\pi r^{2}}{2}=8r-2r^{2}-\pi r^{2}+\frac{\pi r^{2}}{2}=8r-2r^{2}-\frac{\pi r^{2}}{2}$. Domain: $r>0$, and $h>0$ requires $8-2r-\pi r>0$, that is, $r<\frac{8}{2+\pi}\approx 1.556$. The domain is therefore $\left(0,\frac{8}{2+\pi}\right)$.

$\mathcal{A}'(r)=8-4r-\pi r=8-r(4+\pi)$, which vanishes for $r=\frac{8}{4+\pi}\approx 1.1202$ m. It is indeed a maximum, since $\mathcal{A}''(r)=-(4+\pi)<0$: the function is concave down, so its only critical point is an absolute maximum. Corresponding height: $h=\frac{8-2(1.1202)-\pi(1.1202)}{2}\approx 1.1202$ m. Maximum area: $\mathcal{A}\approx 8(1.1202)-2(1.2548)-\frac{\pi(1.2548)}{2}\approx 8.962-2.510-1.971\approx 4.481$ m².

Let us prove it exactly. We have $r=\frac{8}{4+\pi}$, so $8=r(4+\pi)$. Substituting into the expression for $h$: $h=\frac{8-2r-\pi r}{2}=\frac{r(4+\pi)-2r-\pi r}{2}=\frac{4r+\pi r-2r-\pi r}{2}=\frac{2r}{2}=r$ ✓. The relation is therefore EXACT, all the terms in $\pi$ cancelling out. Geometric meaning: the height of the rectangle equals the radius of the semicircle, so it is HALF the width $2r$. The optimal rectangle is thus twice as wide as it is tall, and the whole window fits exactly into half a square topped by its semicircle. It is the same kind of remarkable proportion as the $h=2r$ of the can: geometric optimization problems almost always produce a simple ratio, independent of the numerical data, and checking it symbolically is the best control of the solution.

Marginal cost: $C'(x)=40+0.1x$ dollars per unit. At $x=100$: $C'(100)=40+10=50$ dollars. Concrete interpretation: producing the 101st unit costs about 50 dollars more. Direct check: $C(101)-C(100)=(2000+4040+0.05\times 10201)-(2000+4000+500)=6550.05-6500=50.05$ dollars, which confirms the approximation to within five cents. The marginal cost is thus a differential in disguise: it is $dC$ for $dx=1$, and the 5-cent gap is exactly the neglected quadratic term.

Revenue: $R(x)=x\,p(x)=x(120-0.15x)=120x-0.15x^{2}$. Profit: $P(x)=R(x)-C(x)=120x-0.15x^{2}-2000-40x-0.05x^{2}=-0.2x^{2}+80x-2000$. The profit is a parabola that opens downward, which guarantees from the start that a unique maximum exists.

$P'(x)=-0.4x+80$, which vanishes for $x=200$ units. Since $P''(x)=-0.4<0$, it is indeed a maximum, and it is absolute since the parabola is concave down. Maximum profit: $P(200)=-0.2(40{,}000)+16{,}000-2000=-8000+16{,}000-2000=6000$ dollars. Checking the rule: the marginal revenue is $R'(x)=120-0.3x$, so $R'(200)=120-60=60$ dollars; the marginal cost is $C'(200)=40+20=60$ dollars. They are equal ✓. The rule is proved in one line: $P'=R'-C'$, so setting $P'$ to zero is exactly the same as writing $R'=C'$. The economic interpretation is crystal clear: as long as the next unit brings in more than it costs, keep producing; stop when the two become equal.

$\bar{C}(x)=\frac{2000}{x}+40+0.05x$. Differentiating: $\bar{C}'(x)=-\frac{2000}{x^{2}}+0.05$, which vanishes when $x^{2}=40{,}000$, that is, $x=200$ units. It is a minimum, since $\bar{C}''(x)=\frac{4000}{x^{3}}>0$. Value: $\bar{C}(200)=10+40+10=60$ dollars per unit, and the marginal cost $C'(200)=40+20=60$ dollars: the two coincide ✓. It is NOT a coincidence, and the general proof is short: $\bar{C}(x)=\frac{C(x)}{x}$ gives, by the quotient rule, $\bar{C}'(x)=\frac{C'(x)x-C(x)}{x^{2}}$. Setting this derivative to zero amounts to $C'(x)x=C(x)$, that is, $C'(x)=\frac{C(x)}{x}=\bar{C}(x)$. At the minimum of the average cost, marginal cost and average cost are therefore always equal, whatever the cost function. It is a standard result of microeconomics, and it comes entirely out of the quotient rule. A small coincidence specific to this problem: the production level that minimizes the average cost is here the same as the one that maximizes profit, which is not true in general and comes from the choice of coefficients.

The path on the sand goes from the lifeguard to the entry point: its length is $\sqrt{40^{2}+x^{2}}=\sqrt{1600+x^{2}}$, covered at 8 m/s. The path in the water goes from the entry point to the swimmer: its length is $\sqrt{30^{2}+(100-x)^{2}}=\sqrt{900+(100-x)^{2}}$, covered at 2 m/s. The total time is therefore $T(x)=\frac{\sqrt{1600+x^{2}}}{8}+\frac{\sqrt{900+(100-x)^{2}}}{2}$, on the domain $\left[0, 100\right]$.

$T'(x)=\frac{x}{8\sqrt{1600+x^{2}}}-\frac{100-x}{2\sqrt{900+(100-x)^{2}}}$. Setting it to zero: $\frac{x}{8\sqrt{1600+x^{2}}}=\frac{100-x}{2\sqrt{900+(100-x)^{2}}}$. Interpretation: in each term, the quotient $\frac{x}{\sqrt{1600+x^{2}}}$ is the SINE of the angle the path makes with the normal to the shoreline, and the remaining denominator is the speed. The equation therefore says that the ratio “sine over speed” must be the same on both sides, which part d) makes formal.

The equation cannot be solved algebraically in a simple way (it leads to a polynomial of degree 4), so we solve it numerically, by bisection or by Newton's method. The critical number is $x\approx 92.92$ m, and the corresponding time is $T(92.92)=\frac{\sqrt{1600+8634}}{8}+\frac{\sqrt{900+50.1}}{2}\approx 12.65+15.41\approx 28.06$ s. Justification: on the closed interval $\left[0, 100\right]$, we compare with the endpoints, $T(0)=5+52.20=57.20$ s (entering the water right away) and $T(100)=13.46+15=28.46$ s (running all the way to the perpendicular). Since $28.06<28.46<57.20$, the absolute MINIMUM is indeed at the critical number ✓. These three values outline the graph of $T$: it falls steeply, bottoms out a little before $x=100$, then rises slightly. The result makes physical sense: since swimming is four times slower than running, the lifeguard should run almost the whole distance ($92.9$ m out of 100) before entering the water. Note that the gain over the naive solution “run all the way” is only $0.4$ s: the minimum is very flat, which is typical of optima and explains why an error of a few metres on the entry point costs almost nothing.

Let $\theta_{1}$ be the angle between the path on the sand and the normal to the shoreline, and $\theta_{2}$ the corresponding angle in the water. By the definition of the sine in each right triangle: $\sin\theta_{1}=\frac{x}{\sqrt{1600+x^{2}}}$ and $\sin\theta_{2}=\frac{100-x}{\sqrt{900+(100-x)^{2}}}$. The condition $T'(x)=0$ from part b) then reads exactly $\frac{\sin\theta_{1}}{8}=\frac{\sin\theta_{2}}{2}$, that is, $\frac{\sin\theta_{1}}{v_{1}}=\frac{\sin\theta_{2}}{v_{2}}$ ✓. This is SNELL'S LAW of refraction, obtained here without any physics, by differential calculus alone. Numerical check at the optimum: $\sin\theta_{1}=0.9185$ and $\sin\theta_{2}=0.2296$, hence $\frac{0.9185}{8}=0.11481$ and $\frac{0.2296}{2}=0.11481$ ✓, a perfect match. Setting $n=\frac{c}{v}$ for the index of refraction, the relation takes its usual form $n_{1}\sin\theta_{1}=n_{2}\sin\theta_{2}$. What the result says about light is Fermat's principle: between two points, light follows the path that minimizes the travel time. Light obviously “chooses” nothing, but everything happens as if it were solving the lifeguard's problem. Along the way we recover the behaviour of the denser medium: where the speed is low (water, glass), the path gets shorter and the ray bends toward the normal, exactly as the lifeguard shortens the swimming part.

Let $(x,x^{2})$ be any point on the parabola. Its distance to $A(0, 2)$ is $d=\sqrt{(x-0)^{2}+(x^{2}-2)^{2}}$. We minimize the square instead: $d^{2}(x)=x^{2}+(x^{2}-2)^{2}=x^{2}+x^{4}-4x^{2}+4=x^{4}-3x^{2}+4$. Differentiating: $\left(d^{2}\right)'=4x^{3}-6x=2x\left(2x^{2}-3\right)$, which vanishes for $x=0$ and $x^{2}=\frac{3}{2}$, that is, $x=\pm\sqrt{1.5}\approx\pm 1.2247$. Values: at $x=0$, $d^{2}=4$ so $d=2$; at $x^{2}=1.5$, $d^{2}=2.25-4.5+4=1.75$ so $d=\sqrt{1.75}\approx 1.3229$. The minimum is therefore reached at the TWO points $\left(\pm 1.2247,\, 1.5\right)$, at a distance of $\approx 1.32$ from $A$. The symmetry was predictable, since $A$ lies on the axis of the parabola. The point $x=0$, for its part, is a LOCAL maximum of the distance: it is the vertex of the parabola, farther from $A$ than its neighbours.

It is legitimate because the square root function is strictly INCREASING on $\left[0, +\infty\right)$: if $d^{2}$ reaches its minimum at $x_{0}$, then $d=\sqrt{d^{2}}$ reaches its minimum at the same $x_{0}$. Minimizing the square therefore gives exactly the same optimal point; only the VALUE of the minimum changes (take the root at the end if the question asks for the distance). It is much faster because $d^{2}$ is a polynomial, whereas differentiating $d=\sqrt{\ldots}$ would require the chain rule and leave a radical in the denominator to simplify. The trick works for any strictly increasing function applied to the quantity to optimize: you can minimize $\ln f$ instead of $f$ (very useful for products and exponentials), or $f^{3}$ instead of $f$. Beware, on the other hand, of DECREASING functions, which swap minimum and maximum, and of non-monotonic functions, which preserve nothing.

$P(R)=\frac{E^{2}R}{(R+r)^{2}}$. By the quotient rule: $P'(R)=E^{2}\cdot\frac{(R+r)^{2}-R\cdot 2(R+r)}{(R+r)^{4}}=E^{2}\cdot\frac{(R+r)-2R}{(R+r)^{3}}=\frac{E^{2}(r-R)}{(R+r)^{3}}$. On the domain $R>0$, the denominator is positive, so the sign of $P'$ is that of $r-R$: positive for $R<r$, negative for $R>r$. The power increases then decreases, and the maximum is reached for $R=r$, here $R=2$ ohms. Maximum power: $P(2)=\frac{144\times 2}{(4)^{2}}=\frac{288}{16}=18$ W. We can also write the general formula $P_{\max}=\frac{E^{2}}{4r}=\frac{144}{8}=18$ W ✓. This result has a name in electricity, the maximum power transfer theorem, and it says that a source delivers the most power when the load EQUALS its internal resistance.

At maximum power, the external and internal resistances are equal, so the current flows through two identical resistances and the power is shared equally between them. Explicit computation: the current is $I=\frac{E}{R+r}=\frac{12}{4}=3$ A, the total power supplied by the battery is $EI=36$ W, of which 18 W go to the load and 18 W are dissipated in the battery itself. The efficiency is therefore exactly 50 percent, and this figure depends neither on $E$ nor on $r$. The trade-off for the engineer is real: maximizing the POWER transferred means wasting half the energy as internal heat. That is acceptable when the available power matters more than the energy consumed, for example for an audio signal or an antenna, where one speaks of impedance matching. It is unacceptable, however, for a power distribution grid, where EFFICIENCY is the goal and one therefore imposes $R\gg r$: one then accepts delivering less than the maximum possible power, in exchange for reduced line losses. Maximizing power and maximizing efficiency are two DIFFERENT optimization problems, and confusing them is the classic error on this topic.

If each order is for $q$ units and 10,000 are needed per year, the number of orders per year is $\frac{10{,}000}{q}$, hence an ordering cost of $\frac{10{,}000}{q}\times 50=\frac{500{,}000}{q}$ dollars. The stock decreases linearly from $q$ to 0 between two deliveries, so the AVERAGE stock is $\frac{q}{2}$, hence a holding cost of $\frac{q}{2}\times 4=2q$ dollars. Total annual cost: $C(q)=\frac{500{,}000}{q}+2q$, on the domain $q>0$.

$C'(q)=-\frac{500{,}000}{q^{2}}+2$, which vanishes when $q^{2}=250{,}000$, that is, $q=500$ units (the negative root is excluded from the domain). Since the domain $\left(0, +\infty\right)$ is OPEN, the Extreme Value Theorem does not apply and the justification must come from elsewhere: $C''(q)=\frac{1{,}000{,}000}{q^{3}}>0$ for every $q>0$, so $C$ is CONCAVE UP on the whole domain, and its only critical point is an absolute minimum ✓. One can also observe that $C'<0$ before 500 and $C'>0$ after, or that $C\to+\infty$ at both ends of the domain.

Minimum cost: $C(500)=\frac{500{,}000}{500}+2(500)=1000+1000=2000$ dollars per year. Breakdown: the ORDERING cost is 1000 dollars and the HOLDING cost is also 1000 dollars. The two components are EXACTLY EQUAL at the optimum. Neighbourhood check: $C(400)=1250+800=2050$ and $C(600)\approx 833+1200=2033$, both greater than 2000 ✓.

Let us write the general model with $D$ the annual demand, $K$ the cost per order and $h$ the unit holding cost: $C(q)=\frac{DK}{q}+\frac{hq}{2}$. Then $C'(q)=-\frac{DK}{q^{2}}+\frac{h}{2}=0$ gives $q^{*}=\sqrt{\frac{2DK}{h}}$, known as the Wilson formula (the EOQ formula). At this optimum, the ordering cost is $\frac{DK}{q^{*}}$ and the holding cost $\frac{hq^{*}}{2}$; now the condition $C'(q^{*})=0$ reads precisely $\frac{DK}{\left(q^{*}\right)^{2}}=\frac{h}{2}$, which gives, multiplying both sides by $q^{*}$: $\frac{DK}{q^{*}}=\frac{hq^{*}}{2}$ ✓. The two costs are therefore always equal at the optimum, whatever $D$, $K$ and $h$. MANAGEMENT RULE: a manager does not need to differentiate anything; it is enough to compare the two cost items. If more is spent on ordering fees than on storage, orders are placed too often and the lot size should increase, and vice versa. The equality of the two items is the signal of the optimum, and it is a criterion that can be read directly in the accounts.