Session 05 - Single Variable Optimization

Mathematics for Master Students

Author

Dr. Tobias Vlcek

Recap & Your Questions

Session 04 in a Nutshell

The derivative \(f'(a)\) is the slope of the tangent, the instantaneous rate of change at \(a\)
Power, sum, product, quotient and chain rule differentiate every gallery function
\(e^x\) is its own derivative; \(\ln x\) has derivative \(\frac{1}{x}\)
\(f''\) measures curvature: \(f'' > 0\) convex, \(f'' < 0\) concave, the chord intuition, formalised
Marginal cost \(C'(q) \approx\) cost of one more unit; where \(R'(q) = C'(q)\) hides the best output
Today: we turn a flat tangent into a recipe for finding the best decision

Common Issues in the Homework

The inner derivative went missing: with \(V(t) = 200e^{0.1t}\), the chain rule gives \(V'(t) = 20e^{0.1t}\), the factor \(0.1\) is not optional, so the answer is not \(200e^{0.1t}\)
Quotient order flipped: in \(\left(\frac{\ln x}{x^2}\right)'\) the numerator is \(f'g - fg'\), the minus acts on the second product; swapping the terms flips the sign
A rate read as a total: \(C'(100) = 35\) means the 101st unit costs about €35, not that 100 units cost €35 (that total is \(C(100) = 7500\), i.e. €7500)

. . .

Which tasks gave you trouble? Bring them up now. We take the time to go through them before we move on to today’s topic.

Warm-Up: Find the Flat Spot

Question

Differentiate \(f(x) = 2x^2 - 8x + 3\). At which \(x\) is the tangent horizontal?

. . .

Differentiate term by term: \(\;f'(x) = 4x - 8\)
Horizontal tangent means slope zero: \(\;4x - 8 = 0 \;\Rightarrow\; x = 2\)
At \(x = 2\) the parabola turns, its lowest point, the vertex
That flat spot is exactly where today’s hunt for optima begins

Today’s Plan

Stationary points, a flat tangent flags a candidate: solve \(f'(x) = 0\)
The second-order condition: does \(f''\) say peak or valley?
Local vs global: closed intervals and business boundaries
Optimisation in business: profit maximisation and the economic order quantity

Stationary Points & the First-Order Condition

What Does “Optimal” Mean?

Optimisation means finding the input \(x\) that makes \(f(x)\) as large or as small as possible.

Local maximum at \(x^*\): \(f(x^*) \geq f(x)\) for all nearby \(x\), a peak
Local minimum at \(x^*\): \(f(x^*) \leq f(x)\) for all nearby \(x\), a valley
Global (absolute) extremum: the largest or smallest value on the whole domain
Business speaks this language daily: most profit, least cost, best order size

Where Can an Optimum Hide?

An extremum of a smooth function can only sit in one of three kinds of place:

A stationary point, where \(f'(x) = 0\) (the tangent is flat)
A boundary of the domain, the edge of what is allowed
A kink, where \(f'\) does not exist (Session 4’s non-differentiable points)

. . .

For the smooth business functions in this tutorial, the first two cover almost everything, and stationary points come first.

Your Turn: Where Does It Hide?

Question

\(f(x) = |x - 3|\) has its minimum at \(x = 3\). Which kind of place is that?

(a) A stationary point	(b) A boundary of the domain
(c) A kink	(d) It has no minimum

. . .

(c): the graph is V-shaped, and at the tip no tangent exists, so \(f'(3)\) is undefined and the FOC cannot find it.

. . .

Session 4’s non-differentiable points are not exotic: tiered tariffs and absolute deviations (forecast errors!) produce kinks all the time.

The Flat-Tangent Idea

. . .

At a smooth peak or valley the graph momentarily stops rising or falling, the tangent is horizontal, so its slope \(f'(x)\) is zero.

The First-Order Condition

A point \(x^*\) with \(f'(x^*) = 0\) is called a stationary point (or critical point).

\[f'(x^*) = 0 \qquad \text{the first-order condition (FOC)}\]

It is a necessary condition: every smooth interior extremum must be stationary
So the FOC gives us the candidates, solve \(f'(x) = 0\) to find them all
It is not sufficient: a flat tangent alone does not prove a peak or a valley
Second step (next section) sorts the candidates into maxima, minima and neither

Finding the Candidates

Take \(f(x) = x^3 - 3x^2 - 9x + 5\). Where are its stationary points?

Differentiate: \(f'(x) = 3x^2 - 6x - 9\)
Factor: \(f'(x) = 3(x^2 - 2x - 3) = 3(x - 3)(x + 1)\)
Set to zero: \(3(x - 3)(x + 1) = 0 \;\Rightarrow\; x = -1 \ \text{ or } \ x = 3\)
Two candidates: \(x = -1\) and \(x = 3\), but which is a peak, which a valley?

. . .

Always factor \(f'(x)\) if you can, the stationary points are just its roots, and a factored form hands them to you directly.

Your Turn: Find the Stationary Points

Question

Where are the stationary points of \(\;f(x) = x^3 - 12x\;\)? Shout all of them.

. . .

\[f'(x) = 3x^2 - 12 = 3(x^2 - 4) = 3(x - 2)(x + 2) = 0 \quad \Rightarrow \quad x = 2 \ \text{ or } \ x = -2\]

. . .

Common Mistake

Solving \(x^2 = 4\) as “\(x = 2\)” loses half the candidates. A square root has two signs; keep them both until the second-order test sorts them.

Marginal Revenue Meets Marginal Cost

Session 4 promised: the best output is where marginal revenue meets marginal cost. Here is why.

Profit is revenue minus cost: \(\;\pi(q) = R(q) - C(q)\).

. . .

The first-order condition \(\pi'(q) = 0\) becomes:

\[\pi'(q) = R'(q) - C'(q) = 0 \quad \Longleftrightarrow \quad R'(q) = C'(q)\]

. . .

In words: keep producing while the next unit earns more than it costs
Stop where the extra revenue exactly equals the extra cost, \(R' = C'\)
The abstract “\(f'(x) = 0\)” is the business rule “marginal revenue = marginal cost”

Your Turn: Where Do They Meet?

Question

Marginal revenue is \(R'(q) = 80 - 2q\), marginal cost is \(C'(q) = 20 + q\). At which output is profit maximal? Shout the number.

. . .

\[80 - 2q = 20 + q \quad \Rightarrow \quad 60 = 3q \quad \Rightarrow \quad q = 20\]

. . .

Below \(20\) the next unit earns more than it costs, above \(20\) it costs more than it earns: the gap between revenue and cost is widest at \(q = 20\).

Your Turn: Is a Flat Tangent Enough?

Question

A function satisfies \(f'(2) = 0\). Does that alone tell you \(x = 2\) is a maximum or a minimum?

. . .

No, \(f'(2) = 0\) only makes \(x = 2\) a candidate
The tangent is flat at peaks, at valleys, and at saddle points like \(f(x) = x^3\) at \(0\)
The FOC narrows the search; deciding the type needs a second test, coming up next

Classifying Candidates: the Second-Order Condition

Peak or Valley? The Second Derivative Decides

At a stationary point \(x^*\), curvature settles the question, Session 4’s \(f''\) returns:

\(f''(x^*) > 0\): graph is convex (bends upward), the valley of a bowl → local minimum
\(f''(x^*) < 0\): graph is concave (bends downward), the top of a hill → local maximum
\(f''(x^*) = 0\): the test is inconclusive, could be either, or neither
The chord picture, made into a test: convex holds water, concave spills it

The Recipe in Four Steps

Every optimisation problem in this tutorial runs through the same four steps:

Differentiate: compute \(f'(x)\)
Solve \(f'(x) = 0\): the stationary points are the candidates (FOC)
Differentiate again and evaluate \(f''\) at each candidate
Classify: \(f'' > 0\) minimum, \(f'' < 0\) maximum, \(f'' = 0\) undecided (SOC)

. . .

Say the goal out loud first, maximise or minimise, so you know which sign of \(f''\) you are hoping for.

Classifying Our Candidates

\(f''(x) = 6x - 6\). At \(x = -1\): \(f''(-1) = -12 < 0\) → local maximum, value \(f(-1) = 10\)
At \(x = 3\): \(f''(3) = 12 > 0\) → local minimum, value \(f(3) = -22\)
Where \(f'' = 0\), at \(x = 1\), the bend switches: an inflection point \((1, -6)\)

Your Turn: Classify the Candidates

Question

Back to \(f(x) = x^3 - 12x\) with stationary points \(x = 2\) and \(x = -2\). Which is which?

(a) Maximum at \(2\), minimum at \(-2\)	(b) Minimum at \(2\), maximum at \(-2\)
(c) Both are minima	(d) The test is inconclusive

. . .

(b): \(f''(x) = 6x\), so \(f''(2) = 12 > 0\) (valley) and \(f''(-2) = -12 < 0\) (peak).

. . .

Values: \(f(2) = 8 - 24 = -16\) and \(f(-2) = -8 + 24 = 16\), a local minimum of \(-16\) and a local maximum of \(16\).

When the Test Is Silent: \(f'' = 0\)

. . .

For \(f(x) = x^3\): \(f'(0) = 0\) and \(f''(0) = 0\). The tangent is flat, yet the graph keeps rising, a saddle point, neither peak nor valley.

The FOC Is Never the Whole Story

Common Mistake

Declaring “\(f'(x^*) = 0\), therefore it is a maximum”. A stationary point is only a candidate. Always check the second-order condition (or watch how \(f'\) changes sign across \(x^*\)) before naming it a max or a min.

. . .

Alternative, the first-derivative test: if \(f'\) goes \(+ \to -\) at \(x^*\), it is a maximum; \(- \to +\), a minimum
If \(f'\) keeps its sign across \(x^*\) (as for \(x^3\)), it is neither
The first-derivative test always works, handy exactly when \(f'' = 0\) falls silent

The First-Derivative Test as a Table

For \(f(x) = x^3 - 3x^2 - 9x + 5\), \(f'(x) = 3(x - 3)(x + 1)\). One test point per interval:

Interval	\(x < -1\)	\(-1 < x < 3\)	\(x > 3\)
test point	\(x = -2\)	\(x = 0\)	\(x = 4\)
\(f'(x)\)	\(3 \cdot (-5) \cdot (-1) = 15\)	\(3 \cdot (-3) \cdot 1 = -9\)	\(3 \cdot 1 \cdot 5 = 15\)
sign	\(+\)	\(-\)	\(+\)

. . .

At \(x = -1\) the sign flips \(+ \to -\): rising, then falling, a maximum
At \(x = 3\) it flips \(- \to +\): falling, then rising, a minimum
Same verdict as \(f''\), and it also works when \(f'' = 0\)

Your Turn: Max or Min?

Question

At a stationary point \(x^*\) of a cost function, you compute \(f''(x^*) = 8 > 0\). Is \(x^*\) a local maximum or a local minimum?

. . .

\(f''(x^*) > 0\) means the graph is convex there, a valley
So \(x^*\) is a local minimum, exactly what you want when minimising cost
Mnemonic: positive second derivative → holds water → minimum

Local vs Global & Boundaries

Local Is Not Global

A local optimum is best only in its neighbourhood, a bigger peak may sit elsewhere.

The cubic’s local min at \((3, -22)\) is a valley, but \(f\) dives far lower as \(x \to -\infty\)
On an unbounded domain, a global optimum may not exist at all
Extreme Value Theorem: a continuous function on a closed interval \([a, b]\) always attains both a global max and a global min
Business is usually bounded (quantities live in \([0, \text{capacity}]\)) so globals exist

The Closed-Interval Recipe

To find the global max and min of \(f\) on \([a, b]\), compare a short list:

Find every stationary point in \((a, b)\): solve \(f'(x) = 0\)
Evaluate \(f\) at each of them
Evaluate \(f\) at both endpoints, \(f(a)\) and \(f(b)\)
The largest value is the global max; the smallest is the global min

. . .

Common Mistake

Finding stationary points but forgetting the endpoints. The global optimum often sits at a boundary, skip it and you get the wrong answer.

A Boundary Can Win

Comparing the Candidates

Maximise \(g(x) = x^3 - 3x\) on \([0,3]\):

\(x\)	\(0\) (endpoint)	\(1\) (stationary)	\(3\) (endpoint)
\(g(x)\)	\(0\)	\(-2\)	\(\mathbf{18}\)

. . .

The only interior stationary point is a minimum; the global max \(18\) sits at the boundary \(x = 3\).

Your Turn: Global on an Interval

Question

Find the global maximum of \(\;f(x) = x^2 - 4x\;\) on \([0, 5]\). Shout the maximum value.

. . .

Stationary point: \(f'(x) = 2x - 4 = 0 \Rightarrow x = 2\), with \(f(2) = 4 - 8 = -4\)
Endpoints: \(f(0) = 0\) and \(f(5) = 25 - 20 = 5\)
Compare \(\{-4, 0, 5\}\): the global maximum is \(5\), at the endpoint \(x = 5\); the stationary point is the global minimum

. . .

Common Mistake

Reporting \(-4\) because “that is where \(f' = 0\)”. An upward parabola has no interior maximum at all: on a closed interval its maximum always sits at an endpoint.

Business Domains Have Edges

Real decisions come with hard limits, and those edges are candidates:

Quantities are non-negative: \(q \geq 0\), the edge \(q = 0\) may be optimal (produce nothing)
Capacity caps output: \(q \leq 500\), the unconstrained optimum may lie past the limit
When it does, the best feasible choice is the capacity boundary itself
Model domain = mathematical domain \(\cap\) business constraints, an intersection, as in Session 2

Your Turn: The Capacity Limit

Question

You maximise profit and the FOC gives \(q = 45\). But your plant can make at most \(40\) units, so \(q \in [0, 40]\) and profit is still rising at \(q = 40\). What is the best feasible output?

. . .

The unconstrained optimum \(q = 45\) lies outside the feasible interval
Profit increases all the way up to the edge, so the best feasible choice is the boundary \(q = 40\)
Lesson: check the endpoints, a binding capacity limit is your optimum

Optimisation in Business

Setting Up a Profit Problem

We reuse Session 4’s cost function and add a revenue function:

\[C(q) = 0.5q^2 + 10q + 200 \qquad R(q) = 100q - 0.5q^2\]

\(C\): fixed costs \(200\), rising variable costs, marginal cost \(C'(q) = q + 10\)
\(R\): revenue peaks then falls, since selling more needs a lower price, \(R'(q) = 100 - q\)
Profit is the gap: \(\;\pi(q) = R(q) - C(q)\)
The plan: build \(\pi\), apply the FOC, verify with the SOC

Worked Example: Build the Profit Function

Subtract cost from revenue, term by term:

\[\pi(q) = (100q - 0.5q^2) - (0.5q^2 + 10q + 200)\]

. . .

\[\pi(q) = -q^2 + 90q - 200\]

. . .

A downward parabola, exactly the trade-off shape from Session 3
Differentiate: \(\;\pi'(q) = -2q + 90\)
The fixed cost \(200\) vanishes under differentiation, it shifts profit, not the optimum

Solve and Verify

FOC: \(\pi'(q) = -2q + 90 = 0 \;\Rightarrow\; q = 45\), and \(R'(45) = 55 = C'(45)\), marginal revenue = marginal cost
SOC: \(\pi''(q) = -2 < 0\) → concave → a maximum, confirmed

Your Turn: Fixed Costs Rise

Question

The rent goes up: fixed costs rise from \(200\) to \(500\), so \(C(q) = 0.5q^2 + 10q + 500\). What happens to the optimal output \(q = 45\) and to the maximum profit?

(a) \(q\) falls, profit falls	(b) \(q\) rises, profit falls
(c) \(q\) unchanged, profit falls by \(300\)	(d) \(q\) unchanged, profit unchanged

. . .

(c): \(\pi(q) = -q^2 + 90q - 500\) has the same derivative \(\pi'(q) = -2q + 90\), so the FOC still gives \(q = 45\); the constant only shifts the curve down.

. . .

Fixed costs never change the decision, only the result. That is the practical meaning of “constants differentiate to zero”.

Reading the Optimum

. . .

The optimal quantity is \(q = 45\), and the maximum profit is \[\pi(45) = -45^2 + 90 \cdot 45 - 200 = -2025 + 4050 - 200 = 1825,\] that is €1825, the highest point of the profit curve.

Your Turn: A Better Price

Question

Demand improves and revenue becomes \(R(q) = 120q - 0.5q^2\), cost stays \(C(q) = 0.5q^2 + 10q + 200\). What is the new optimal output? Shout the number.

. . .

Profit: \(\pi(q) = 120q - 0.5q^2 - 0.5q^2 - 10q - 200 = -q^2 + 110q - 200\)
FOC: \(\pi'(q) = -2q + 110 = 0 \Rightarrow q = 55\)
SOC: \(\pi''(q) = -2 < 0\), a maximum; and \(R'(55) = 65 = C'(55)\) ✓

Worked Example: Minimising Cost per Unit

A plant’s total cost is \(C(q) = 400 + 4q + 0.01q^2\). Which output makes the cost per unit smallest?

Average cost: \(A(q) = \dfrac{C(q)}{q} = \dfrac{400}{q} + 4 + 0.01q\)
Two opposing forces: the fixed cost spreads thinner (\(400/q\) falls), the variable cost per unit grows (\(0.01q\) rises)
Rewrite for the power rule: \(A(q) = 400q^{-1} + 4 + 0.01q\)

Solve: Where Cost per Unit Bottoms Out

FOC: \(A'(q) = -\dfrac{400}{q^2} + 0.01 = 0 \;\Rightarrow\; q^2 = 40000 \;\Rightarrow\; q = 200\)
SOC: \(A''(q) = \dfrac{800}{q^3} > 0\) for \(q > 0\): convex, a minimum
Value: \(A(200) = 2 + 4 + 2 = 8\), at \(200\) units each unit costs €8 on average

. . .

Notice the shape \(\frac{a}{q} + bq\): a falling term against a rising one. The next example, the most famous formula in inventory management, has exactly the same shape.

Logistics: The Economic Order Quantity

A warehouse orders a product with steady annual demand. Order too often and ordering costs pile up; order too much and holding costs pile up. Balance them.

Annual demand \(D\), fixed cost \(K\) per order, holding cost \(h\) per unit per year, order size \(q\)
Ordering cost: \(\frac{D}{q} \cdot K\), \(\frac{D}{q}\) orders per year, each costing \(K\)
Holding cost: \(\frac{q}{2} \cdot h\), average stock is \(\frac{q}{2}\) (fill to \(q\), draw down to \(0\))
Total: \(\;T(q) = \dfrac{D}{q}\,K + \dfrac{q}{2}\,h\)

Your Turn: Bigger Batches

Question

You switch to larger orders \(q\), same annual demand \(D\). What happens to the two cost terms in \(T(q) = \frac{D}{q}K + \frac{q}{2}h\)?

(a) Ordering cost rises, holding cost falls	(b) Ordering cost falls, holding cost rises
(c) Both fall	(d) Both rise

. . .

(b): fewer orders per year (\(D/q\) shrinks), but a higher average stock (\(q/2\) grows).

. . .

Exactly this tug of war makes the total cost U-shaped, and a U has a bottom to find.

The EOQ: First-Order Condition

With \(D = 1000\), \(K = 80\), \(h = 4\), so \(T(q) = 80000/q + 2q\):

FOC: \(T'(q) = -80000/q^2 + 2 = 0 \;\Rightarrow\; q^2 = 40000 \;\Rightarrow\; q^* = 200\)

The EOQ: Verify and Read

SOC: \(T''(q) = 160000/q^3 > 0\) for all \(q > 0\) → convex → a minimum
At \(q^* = 200\): ordering cost \(= 400\) and holding cost \(= 400\), equal, total €800

The Square-Root Formula

Solve the FOC in symbols and the classic EOQ formula drops out:

\[\frac{D K}{q^2} = \frac{h}{2} \;\Rightarrow\; q^2 = \frac{2 D K}{h} \;\Rightarrow\; \boxed{\,q^* = \sqrt{\dfrac{2 D K}{h}}\,}\]

. . .

Check: \(\sqrt{\frac{2 \cdot 1000 \cdot 80}{4}} = \sqrt{40000} = 200\) ✓
The optimum always balances the two costs, that is why they came out equal
One tidy formula behind textbook inventory management, derived in four lines

Your Turn: Compute the EOQ

Question

Annual demand \(D = 1800\) units, \(K = 25\) per order, holding cost \(h = 4\) per unit and year. What is the economic order quantity? Shout the number.

. . .

\[q^* = \sqrt{\frac{2 \cdot 1800 \cdot 25}{4}} = \sqrt{\frac{90000}{4}} = \sqrt{22500} = 150\]

. . .

Check the balance: ordering cost \(\frac{1800}{150} \cdot 25 = 300\), holding cost \(\frac{150}{2} \cdot 4 = 300\), equal, as the theory promised.

From Order Size to Order Frequency

The EOQ also fixes the rhythm of ordering:

Orders per year: \(\dfrac{D}{q^*} = \dfrac{1800}{150} = 12\), one order a month
Time between orders: \(\dfrac{q^*}{D}\) years \(= \dfrac{150}{1800} = \dfrac{1}{12}\) year, about \(30\) days
In our first example (\(D = 1000\), \(q^* = 200\)): \(5\) orders a year, one every \(73\) days

. . .

The formula answers “how much”; dividing by demand answers “how often”. Both are needed for a replenishment plan.

Your Turn: Ordering Cost Doubles

Question

A supplier doubles the fixed cost \(K\) per order. According to \(q^* = \sqrt{\frac{2DK}{h}}\), does the economic order quantity rise or fall, and by how much?

. . .

\(q^*\) grows with \(K\) under the root, so a larger \(K\) makes \(q^*\) rise
Doubling \(K\) multiplies \(q^*\) by \(\sqrt{2} \approx 1.41\), about a 41% larger order
Intuition: if each order costs more, you order less often in bigger batches

Closing

Key Takeaways

First-order condition: optima hide where \(f'(x) = 0\), solve it for the candidates
Second-order condition: \(f''(x^*) > 0\) a minimum, \(f''(x^*) < 0\) a maximum, \(f'' = 0\) inconclusive, the FOC alone is never enough, since a flat tangent can be a saddle (\(x^3\) at \(0\))
On a closed interval, compare stationary points and endpoints, boundaries can win
Profit is maximal where marginal revenue = marginal cost, Session 4’s promise, delivered
The EOQ \(q^* = \sqrt{2DK/h}\) balances ordering against holding costs

Skip order if running long: 1. Reading the Optimum plot (state \(\pi(45)=1825\) verbally on the previous slide), 2. The first-derivative-test bullets on the “FOC Is Never the Whole Story” slide, 3. Business Domains Have Edges (fold into the Your Turn capacity slide), 4. compress the R & C plot to a single sentence about the widest gap.

That’s it for today.

Any Questions?

Until the Next Session

Work through the Tasks: optimisation drills with worked solutions
Check yourself with the Self-Test quiz
Keep the Cheatsheet next to you, the FOC/SOC recipe lives on it
Note down anything unclear, we start next session with your questions

. . .

The recipe is always the same: differentiate, set \(f' = 0\), solve, then let \(f''\) decide. Name the goal (maximise or minimise) and the sign of \(f''\) you want follows.

Preview: Session 06: Systems of Linear Equations

Central question: solve for several unknowns at once
From one equation to a whole system, where do the lines meet?
Elimination and substitution, done systematically
Gaussian elimination for bigger systems, market equilibrium, production planning

. . .

See you there, and bring your questions!

Literature & Further Reading

These sessions cover the essentials: textbooks offer more depth and practice
Start with Sydsæter et al. (2012) or Jacques (2015); full recommendations on the tutorial’s literature page

Jacques, Ian. 2015. Mathematics for Economics and Business. 8th ed. Always Learning. Pearson.

Sydsæter, Knut, Peter J. Hammond, and Arne Strøm. 2012. Essential Mathematics for Economic Analysis. 4th ed. Pearson.