Optimization — constrained, unconstrained, real-world problems: Question bank

Definition Vocabulary refresher (so no symbol is unearned)

Critical point $c$ : a point where $f^{'} (c) = 0$ or $f^{'} (c)$ does not exist. Think "the graph goes flat or has a sharp corner/spike."
Interior point: a point strictly inside the allowed interval, not an endpoint. On $[a, b]$ the interior is $a < x < b$ .
Local extremum: highest/lowest compared to nearby points only. Global (absolute) extremum: highest/lowest over the whole domain.
Increment $h$ : a tiny step we take away from a point $c$ , landing at $c + h$ . It can be positive (step right) or negative (step left); we compare $f (c + h)$ with $f (c)$ to see whether $f$ rose or fell.
Twice differentiable near $c$ : both $f^{'}$ and $f^{''}$ exist for every $x$ in a small open interval around $c$ (a neighbourhood). This is the fine print the second-derivative test needs.
$f^{'}$ (first derivative) = the slope = how fast $f$ climbs. $f^{''}$ (second derivative) = how the slope itself changes = the bend (concavity).
Constraint function $g$ : the side-condition, written as $g (x, y) = 0$ (e.g. " $x + y = 10$ " becomes $g = x + y - 10$ ). It carves out the curve you are forced to walk along.
$\nabla f$ ("grad f") = the arrow pointing in the direction $f$ increases fastest; its length is the steepness. $\nabla g$ = the same arrow for the constraint function; it points perpendicular to the constraint curve $g = 0$ . $λ$ (lambda) = the Lagrange multiplier, a single number linking those two gradient arrows.
Supremum / infimum: the least upper bound / greatest lower bound of a set of values — the tightest ceiling/floor the values crowd up against. A supremum may exist without ever being reached (see the open-interval trap below).

True or false — justify

Every local maximum of a differentiable function is a critical point.

True. By Fermat's Theorem, a differentiable function cannot have positive slope on one side and negative on the other without passing through slope zero, so

f^{'} = 0

there.

Every critical point of a differentiable function is a local maximum or minimum.

False.

f (x) = x^{3}

has

f^{'} (0) = 0

yet

0

is a flat inflection — the graph keeps rising through it. Critical points are only candidates.

f^{''} (c) > 0

then

c

is automatically a local minimum.

False as stated — you must first know

f^{'} (c) = 0

and that

f

is twice differentiable near

c

. The second-derivative test only classifies critical points;

f^{''} > 0

alone just says the graph is cup-shaped there, which happens on rising slopes too.

A continuous function on a closed interval

[a, b]

always attains a global maximum and minimum.

True (Extreme Value Theorem). Closedness + continuity guarantee the extremes are reached, either at a critical point or an endpoint.

A continuous function on an open interval

(a, b)

always attains a global maximum.

False. On

(0, 1)

f (x) = x

climbs toward

1

but never reaches it —

1

is the supremum (least upper bound) yet is not attained. Openness kills the guarantee.

To find the global extremum on

[a, b]

it is enough to compare all critical points.

False. You must also test the two endpoints

a

and

b

; the global extreme often sits at a boundary where

f^{'}

need not vanish.

Minimizing distance

D

and minimizing

D^{2}

give the same minimizing point.

True. Since

\cdot

is strictly increasing on

[0, \infty)

, wherever

D^{2}

is smallest,

D

is smallest too — but

D^{2}

avoids messy square-root derivatives.

\nabla f = λ \nabla g

has a solution, that point is guaranteed to be the constrained maximum.

False. The Lagrange condition finds candidates — it flags maxima, minima, and saddles alike. You still must compare

f

-values (or check the geometry) to know which one you have.

The Lagrange multiplier

λ

is always positive.

False.

λ

can be positive, negative, or zero; its sign tells you the direction in which loosening the constraint changes the optimum. Its magnitude is the sensitivity.

Spot the error

f (x) = x^{4}

has

f^{'} (0) = 0

and

f^{''} (0) = 0

, so

x = 0

is neither a max nor a min."

Wrong conclusion.

f^{''} (0) = 0

only means the test is inconclusive, not that there's no extremum. The first-derivative sign test shows

f^{'} < 0

left of

0

and

f^{'} > 0

right — so

0

is a genuine minimum.

"In the open-box problem — a

12 \times 12

cm sheet, corners of side

x

cut, giving

V (x) = x (12 - 2 x)^{2}

— I solved

V^{'} (x) = 0

and got

x = 6

, so cut 6 cm squares."

Domain error. With base

12 - 2 x

, the value

x = 6

makes

12 - 2 x = 0

, giving zero volume. It's a legitimate root of

V^{'}

but a minimum at the boundary; the real answer is the interior root

x = 2

(volume

128 cm^{3}

"To maximize

f

[a, b]

I set

f^{'} = 0

; there was no solution, so no maximum exists."

Wrong. If

f^{'}

never vanishes, the function is monotonic, so the maximum simply sits at an endpoint. Absence of interior critical points does not abolish extrema.

f^{'} (c) = 0

, so

c

must be smooth — corners never count."

Backwards. A corner or cusp is a critical point precisely because

f^{'} (c)

does not exist there. The definition includes both "slope zero" and "slope undefined."

"I substituted the constraint into

f

, differentiated, and got a critical point — so I never need to check endpoints of the constraint region."

Error. Substitution turns it into a 1-D problem that still has endpoints (physical limits of the variable). Skipping them can miss the true extreme, exactly as in the box problem.

\nabla f = λ \nabla g

means

f

and

g

are equal at the optimum."

Confusion of quantities. It means the gradient arrows are parallel (one is a scalar multiple of the other), not that the function values match. It's about direction, not size or value.

Why questions

Why does the sign of

f^{''} (c)

decide max vs min, and not its size?

Near a critical point (with

f

twice differentiable there), the increment gives

f (c + h) - f (c) \approx \frac{1}{2} f^{''} (c) h^{2}

; since

h^{2} > 0

for any small step

h \neq = 0

, only the sign of

f^{''} (c)

flips the change up or down. See Taylor Series Expansion and Fig 1.

Why can we ignore the

f^{'} (c) h

term in that expansion?

Because at a critical point

f^{'} (c) = 0

, so that whole term vanishes, leaving the

h^{2}

term as the leading behaviour that governs the local shape.

Why must

\nabla f

be perpendicular to the constraint curve at a constrained optimum?

\nabla f

had any component along the curve, we could slide that way and increase

f

while staying feasible — so no optimum yet. Only when the tangent component is zero (i.e.

\nabla f ∥ \nabla g

) are we stuck. See Lagrange Multipliers (Multivariable) and Fig 3.

Why do we prefer to reduce a constrained problem to one variable before differentiating?

A single-variable function has an ordinary derivative we can set to zero directly, using the same Closed Interval logic; it sidesteps needing full multivariable machinery when the constraint is easy to solve.

Why does "fixed sum, maximum product" force the numbers to be equal?

Lagrange gives

x = y

(both partials of the sum are equal), and geometrically the product

x y

grows as the two factors balance — this is AM-GM Inequality wearing a calculus disguise.

Why check perpendicularity as a sanity test in the closest-point problem?

The shortest segment from a point to a line always meets the line at a right angle; if your answer's segment isn't perpendicular (slopes multiplying to

- 1

), you've made an error. See Fig 2.

Why is optimization deeply related to Related Rates even though one seeks extremes and the other seeks speeds?

Both start by translating a word problem into a symbolic relation and then differentiating — the calculus machinery is identical; only the question asked of

f^{'}

differs.

Edge cases

What if

f^{'}

is zero on a whole flat interval, not just a point?

Every point of that interval is a critical point; the function is constant there, so it is simultaneously a (weak) local max and min. No single "peak" — you report the constant value.

What happens at an endpoint where

f^{'} \neq = 0

but

f

is still largest?

The endpoint is a boundary maximum; Fermat's Theorem does not apply (it's only for interior points), which is exactly why the Closed Interval Method tests endpoints separately.

Degenerate box: what if the sheet side were

0

Then

12 - 2 x \to 0

leaves no material; the domain collapses to the single point

x = 0

with volume

0

. Always confirm the feasible domain is non-empty before optimizing.

Constraint and objective gradients both zero (

\nabla f = \nabla g = 0

) at a point — what does Lagrange say?

The equation

\nabla f = λ \nabla g

is satisfied by any

λ

, so the method gives no information; you must analyze that point directly by inspecting values nearby.

A function has two critical points with the same

f

-value on

[a, b]

— which is the global max?

If both tie and exceed the endpoints, the global maximum is attained at both simultaneously. Global extrema need not be unique.

The second-derivative test says minimum, but the point is at the domain boundary — trust it?

Be careful: the test describes local behaviour of the smooth function, but the boundary may cut off one side. Fall back to comparing actual values (Closed Interval Method) to settle the global picture.

Recall One-line self-test before you close this page

Name the two things a critical point could be besides an extremum, and the one place an extremum can hide that has no critical point. Answer ::: An inflection (like $x^{3}$ at $0$ ) or a flat plateau; and an endpoint of the domain.

Question bank — Optimization — constrained, unconstrained, real-world problems