Mathematics, four units · Grade 11

How do you find the largest area you can fence with a given length of fence?

Write the area as a function of one variable. You start with two — a length and a width — and use the constraint on the total fence to express one in terms of the other. Then differentiate the area function, set it to zero, and confirm that the point you found is genuinely a maximum and not a minimum.

Learning objectives

The step before the derivative

An optimisation problem is a word problem with a derivative in the middle, and nearly all the difficulty lies in the sentence before the derivative. The differentiation itself is the easy part.

Start by asking exactly what is to be maximised or minimised: an area, a volume, a cost, a distance. Give that quantity a name and express it using the variables in the question.

Almost always two variables appear. Then look in the question for the constraint — the given perimeter, the fixed volume, the budget — and use it to eliminate one of them. Only when a single variable remains can you differentiate.

Differentiate, solve, verify

Once you have a function of one variable, differentiate and set it to zero. The solutions are the candidates.

You must verify the point is of the right type. The convenient method is testing the sign of the derivative on both sides: plus to minus is a maximum, which is what a largest-area question wants.

Check the practical domain too. A side length must be positive, and with 40 metres of fence the width cannot exceed 20. A solution outside that range is rejected even if the derivative does vanish there.

Answer in the units of the question

The final answer is not the x value. If the question asked for the maximum area, substitute the x you found back into the area function and answer in square metres.

If it asked for the dimensions, give both: the length and the width. Eliminating the second variable at the start means you have to restore it at the end.

One familiar result worth knowing: for a closed rectangular fence of given length the maximum area is a square. If one side is an existing wall the answer changes — the side parallel to the wall comes out twice the width. Knowing both lets you check yourself at the end.

Worked examples

  1. A rectangle is fenced with 40 metres of fence. What is the largest possible area?

    1. Length plus width is 20, so the width is 20 minus x
    2. Area: S equals x times (20 minus x), that is 20x minus x squared
    3. Derivative: 20 minus 2x, set to zero
    4. x equals 10, and the width is 10 as well

    Answer: 100 square metres, a square of side 10

  2. The same fence, but one side is an existing wall. What is the maximum area?

    1. The fence covers two widths and one length: 2x plus y equals 40
    2. y equals 40 minus 2x
    3. Area: x times (40 minus 2x), that is 40x minus 2x squared
    4. Derivative: 40 minus 4x, so x equals 10 and y equals 20

    Answer: 200 square metres — twice the closed case

  3. An open tank with a square base must hold 32 cubic metres. What dimensions minimise the sheet metal?

    1. Volume: x squared times h equals 32, so h equals 32 over x squared
    2. Sheet area: the base plus four walls, that is x squared plus 4xh
    3. Substitute: x squared plus 128 over x
    4. Derivative: 2x minus 128 over x squared, set to zero

    Answer: x equals 4 and h equals 2

Common mistakes

Differentiating a function of two variables
You cannot differentiate until the constraint has eliminated one of them. That step is the whole difference between an optimisation problem that works out and one that stalls.
Answering with the x value when the area was asked for
x is the dimension that maximises, not the maximum value itself. Substitute it back into the function and answer in the units the question used.
Skipping the check on which kind of extreme it is
A vanishing derivative can give a minimum where a maximum was wanted. A cost problem and an area problem look identical up to this point, and without the check you can answer the opposite of what was asked.

What to remember

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