MISSION · Calculus
Finding Maximums and Minimums
Where the slope is zero, and what to do with that
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The idea in one sentence
At the top of a hill and the bottom of a valley the curve is momentarily flat, so setting the derivative to zero finds every candidate for a maximum or minimum.
The method
- Write the quantity you want to maximise or minimise as a function of one variable.
- Differentiate it.
- Set the derivative to zero and solve. These are the stationary points.
- Decide what kind each one is.
- Check the endpoints too.
Step 1 is the hard one in real problems, and it’s the only step that isn’t mechanical. Usually you start with two variables and a constraint linking them — a fixed perimeter, a fixed budget — and use the constraint to eliminate one.
Telling a maximum from a minimum
A stationary point could be a peak, a trough, or neither. Two ways to tell:
The second derivative. If \(f''(x) < 0\) the curve is bending downwards, so it’s a maximum. If \(f''(x) > 0\) it’s bending upwards, a minimum. The mnemonic that survives exams: a negative second derivative is a frowning curve, a positive one is smiling.
The sign of \(f'\) either side. Slope positive then negative means you went over a peak. Negative then positive means a trough. Slower, but it never fails — and it’s the one to use when \(f'' = 0\), which leaves the second-derivative test silent.
That silent case is a point of inflection: flat, but not a turning point. \(y = x^3\) at the origin is the standard example, level for an instant and then carrying straight on up.
Don’t forget the endpoints
If the variable is restricted to a range — a length can’t be negative, a box can’t be longer than the cardboard — the largest value may occur at the end of that range rather than at a stationary point.
The derivative knows nothing about endpoints; it only finds flat spots in the interior. Evaluate the function at both ends as well as at every stationary point, then compare. This is the most commonly dropped mark in the topic.
A worked shape
The classic: a farmer has 100 m of fence and wants the largest rectangular field against a straight river, so only three sides need fencing.
With width \(x\) and depth \(y\): \(2x + y = 100\), so \(y = 100 - 2x\). The area is
\[A = x(100 - 2x) = 100x - 2x^2\]Differentiate: \(A' = 100 - 4x\). Set to zero: \(x = 25\). Then \(y = 50\), and the area is 1250 m². The second derivative is \(-4\), negative, so it’s a maximum.
Every optimisation problem has this shape: constraint, substitute, differentiate, solve, classify.
Common mix-ups
- Stopping at \(f'(x) = 0\) without saying which kind of point it is.
- Ignoring the endpoints of a restricted range.
- Assuming every stationary point is a turning point. Inflections are flat too.
- Optimising the wrong quantity — differentiating the constraint instead of the thing you want to maximise.
Check yourself
A rectangle has a perimeter of 40. What dimensions maximise its area?
With width \(x\), the height is \(20 - x\), so \(A = 20x - x^2\) and \(A' = 20 - 2x\). That’s zero at \(x = 10\), and \(A'' = -2 < 0\), a maximum. A 10 by 10 square, area 100.
LEVEL 3 · REFLECT
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