How do you know whether a stationary point is a maximum or a minimum?

Use the second-derivative test. After solving f prime of x equals 0, evaluate f double prime at that point: if it is negative the point is a maximum, and if it is positive it is a minimum. A first-derivative sign test - checking the sign of f prime just before and after the point - works too.

Why do you have to check the domain in an optimisation problem?

A solution of f prime of x equals 0 is only a candidate. Physical variables have a restricted domain - a length must be positive, and the square cut from a 12 cm sheet must satisfy 0 less than x less than 6 - so any root outside that range is rejected. On a closed domain you also compare the stationary value with the values at the endpoints.

Should you give the value of x or the value of the quantity as the answer?

Give whatever the question asks for. Sometimes it wants the optimal dimension, the value of x; sometimes it wants the optimal value, f of x, such as the maximum volume. A common error is to stop at x when the question asked for the maximum or minimum value itself.

Do you set the first or the second derivative to zero?

Set the first derivative to zero to locate the stationary point where the maximum or minimum occurs. The second derivative is then evaluated at that point to classify it. Setting f double prime equal to zero instead finds a point of inflection, not the optimum.

How do you know whether a stationary point is a maximum or a minimum?

Use the second-derivative test. After solving f prime of x equals 0, evaluate f double prime at that point: if it is negative the point is a maximum, and if it is positive it is a minimum. A first-derivative sign test - checking the sign of f prime just before and after the point - works too.

Why do you have to check the domain in an optimisation problem?

A solution of f prime of x equals 0 is only a candidate. Physical variables have a restricted domain - a length must be positive, and the square cut from a 12 cm sheet must satisfy 0 less than x less than 6 - so any root outside that range is rejected. On a closed domain you also compare the stationary value with the values at the endpoints.

Should you give the value of x or the value of the quantity as the answer?

Give whatever the question asks for. Sometimes it wants the optimal dimension, the value of x; sometimes it wants the optimal value, f of x, such as the maximum volume. A common error is to stop at x when the question asked for the maximum or minimum value itself.

Do you set the first or the second derivative to zero?

Set the first derivative to zero to locate the stationary point where the maximum or minimum occurs. The second derivative is then evaluated at that point to classify it. Setting f double prime equal to zero instead finds a point of inflection, not the optimum.

Year 12 Methods (Unit 3 & 4) The second derivative and applications

Optimisation problems

Q: What is an optimisation problem?

An optimisation problem asks for the largest or smallest value a quantity can take, such as the maximum area of a paddock or the minimum surface area of a can. You model the quantity as a function of one variable, then use calculus to find its maximum or minimum by locating where the derivative is zero.

Q: How do you set up an optimisation problem with one variable?

Name the quantity to optimise and the variables, then use the constraint given in the problem - a fixed perimeter, area or volume - to write the second variable in terms of the first. Substitute so the quantity depends on a single variable, and note the sensible domain for that variable.

20 practice questions 0 video lessons Theory + worked examples

In Year 12 Mathematical Methods (Queensland, QCAA), optimisation uses calculus to find the largest or smallest value a quantity can take — maximum area or volume, minimum surface area, cost, material or time. Write the quantity as a function of one variable using the constraint, differentiate, solve \(f'(x)=0\), then verify the maximum or minimum with the second-derivative test and check the domain.

An optimisation problem asks for the greatest or least value of some quantity — the maximum area a fixed length of fencing can enclose, the minimum material for a can of given volume, the shortest travel time. The quantity to be optimised depends, at first, on two or more variables, but a constraint in the problem (a fixed perimeter, area or volume) links them.

The key move is to use that constraint to write the quantity as a function of a single variable \(f(x)\). A smooth function reaches a local maximum or minimum only at a stationary point, where \(f'(x)=0\). Solving that equation locates the candidate; the second-derivative test then classifies it — a maximum when \(f''(x)<0\) and a minimum when \(f''(x)>0\).

Two checks separate a full-mark solution from a careless one. First, the variable has a sensible domain (a length is positive), so any root outside it is rejected, and on a closed domain the endpoints are compared too. Second, you must state the quantity the question actually asks for: sometimes the optimal dimension \(x\), sometimes the optimal value \(f(x)\).

Key idea. Write the quantity as \(f(x)\) using the constraint, solve \(f'(x)=0\), classify with \(f''\) (\(<0\) max, \(>0\) min), reject roots outside the domain, and answer what was asked.

Maximum: concave down, \(f''<0\)

Minimum: concave up, \(f''>0\)

Optimise \(f(x)\): the stationary points are the solutions of

\[f'(x)=0\]

f^{'} (x) = 0

The second-derivative test classifies a stationary point \(x=a\):

\[f''(a)<0 \Rightarrow \text{maximum} \qquad f''(a)>0 \Rightarrow \text{minimum}\]

f^{''} (a) < 0

Typical constraints and the quantity built from them (one variable after substitution):

\[\text{rectangle: } 2x+2y=P,\ A=xy \qquad \text{cylinder: } \pi r^{2}h=V,\ S=2\pi r^{2}+2\pi r h\]

A = x y

Verify, do not assume. A root of \(f'(x)=0\) is only a candidate. Confirm the maximum or minimum with \(f''\) (or a first-derivative sign test) and reject any root outside the domain such as \(0
    

How to solve an optimisation problem

Name the quantity and the variables. Draw a diagram; label the quantity to be optimised and the unknown lengths.
Use the constraint to get one variable. The fixed perimeter, area or volume lets you write the second variable in terms of the first, then substitute so the quantity is \(f(x)\). State the domain (e.g. \(0
Differentiate and solve \(f'(x)=0\). Find the stationary point(s).
Classify and check the domain. Use \(f''(x)<0\) (maximum) or \(f''(x)>0\) (minimum); reject roots outside the domain and compare endpoints if the domain is closed.
State the answer asked for. Give the optimal dimension \(x\) or the optimal value \(f(x)\), with units.

Which derivative? Set \(f'(x)=0\) to locate the optimum. Setting \(f''(x)=0\) finds a point of inflection instead — a common slip.

Example 1 — Maximum area

A farmer has \(80\text{ m}\) of fencing for a rectangular paddock. Find the dimensions of greatest area.

Solution

\(2x+2y\)	\(=\)	\(80 \Rightarrow y=40-x\)
\(A\)	\(=\)	\(x(40-x)=40x-x^{2}\)
\(A'\)	\(=\)	\(40-2x=0 \Rightarrow x=20\)
\(A''\)	\(=\)	\(-2<0 \Rightarrow \text{maximum}\)

So \(x=20\), \(y=20\): a \(\mathbf{20\text{ m}\times 20\text{ m}}\) square, area \(400\text{ m}^{2}\).

A = 40 x - x^{2}

Example 2 — Minimum surface area

A closed cylindrical can holds \(500\pi\text{ cm}^{3}\). Find the radius that minimises its surface area.

Solution

\(h\)	\(=\)	\(\dfrac{500}{r^{2}}\ \ (\pi r^{2}h=500\pi)\)
\(S\)	\(=\)	\(2\pi r^{2}+\dfrac{1000\pi}{r}\)
\(S'\)	\(=\)	\(4\pi r-\dfrac{1000\pi}{r^{2}}=0\)
\(r^{3}\)	\(=\)	\(250 \Rightarrow r\approx 6.30\text{ cm}\)

\(S''=4\pi+\dfrac{2000\pi}{r^{3}}>0\), so this is a \(\mathbf{minimum}\).

S = 2 π r^{2} + \frac{1000 π}{r}

Example 3 — Maximum volume (open box)

An open box is made from a \(12\text{ cm}\times 12\text{ cm}\) sheet by cutting a square of side \(x\) from each corner and folding up the flaps. Find the \(x\) that maximises the volume.

Solution

\(V\)	\(=\)	\(x(12-2x)^{2},\ 0
	\(=\)	\(4x^{3}-48x^{2}+144x\)
\(V'\)	\(=\)	\(12(x-2)(x-6)=0\)
\(x\)	\(=\)	\(2 \ (\text{reject } x=6)\)
\(V''(2)\)	\(=\)	\(-48<0 \Rightarrow \text{max}\)

\(x=\mathbf{2\text{ cm}}\), giving a maximum volume \(V=128\text{ cm}^{3}\).

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

Example 4 — Minimum time

A river is \(3\text{ km}\) wide. A rower rows at \(8\text{ km/h}\) to a point \(x\text{ km}\) downstream, then cycles at \(10\text{ km/h}\) the remaining \((6-x)\text{ km}\). Minimise the total time.

Solution

\(T\)	\(=\)	\(\dfrac{\sqrt{x^{2}+9}}{8}+\dfrac{6-x}{10}\)
\(T'\)	\(=\)	\(\dfrac{x}{8\sqrt{x^{2}+9}}-\dfrac{1}{10}=0\)
\(100x^{2}\)	\(=\)	\(64(x^{2}+9)\)
\(x\)	\(=\)	\(4 \ (T''>0:\ \text{minimum})\)

Row to \(x=\mathbf{4\text{ km}}\), then cycle the last \(2\text{ km}\), for the least time.

T = \frac{\sqrt{x^{2} + 9}}{8} + \frac{6 - x}{10}

Common pitfalls

Not verifying the stationary point. A solution of \(f'(x)=0\) is only a candidate. Always confirm it is a maximum (\(f''<0\)) or a minimum (\(f''>0\)) with the second-derivative test or a sign test.

Ignoring the domain. Physical variables are restricted — the corner cut of a \(12\text{ cm}\) sheet satisfies \(0

Answering with the wrong quantity. If the question asks for the maximum volume, give \(f(x)\), not the \(x\) that produces it. And set \(f'(x)=0\), not \(f''(x)=0\), to locate the optimum.

Frequently asked questions

What is an optimisation problem?

It asks for the largest or smallest value a quantity can take — maximum area or volume, minimum surface area, cost or time — found by modelling the quantity as a function of one variable and using calculus.

How do you get down to one variable?

Use the constraint (a fixed perimeter, area or volume) to write the second variable in terms of the first, then substitute so the quantity is \(f(x)\).

How do you tell a maximum from a minimum?

Evaluate the second derivative at the stationary point: \(f''<0\) gives a maximum, \(f''>0\) a minimum. A first-derivative sign test works too.

Why check the domain?

A length must be positive, so a root outside the sensible range is rejected; on a closed interval the endpoints may give the true optimum.

Do you give \(x\) or the value of the quantity?

Whatever the question asks: the optimal dimension \(x\), or the optimal value \(f(x)\) such as the maximum volume, with units.

First or second derivative to zero?

Set \(f'(x)=0\) to locate the optimum; \(f''(x)=0\) finds a point of inflection instead.

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