Stationary points and optimisation flashcards

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Stationary point
A point on a curve where dy/dx = 0, so the tangent is horizontal.
Local maximum
A stationary point where the curve changes from increasing to decreasing (gradient + then −).
Local minimum
A stationary point where the curve changes from decreasing to increasing (gradient − then +).
Stationary point of inflection
A stationary point where the gradient has the same sign on both sides, so the curve flattens but does not turn.
Increasing function (on an interval)
f′(x) > 0 throughout the interval: y gets larger as x gets larger.
Decreasing function (on an interval)
f′(x) < 0 throughout the interval: y gets smaller as x gets larger.
Second derivative d²y/dx²
The derivative of dy/dx: the rate of change of the gradient.
Condition for a stationary point
dy/dx = 0 (No — learn it)
Second derivative test
At a stationary point: d²y/dx² < 0 ⇒ maximum; d²y/dx² > 0 ⇒ minimum; = 0 ⇒ inconclusive. (No — learn it)
Volume and curved surface area of a cylinder
V = πr²h; curved surface 2πrh; each circular end πr². (No — learn it)
Surface area of a closed cuboid l × w × h
2(lw + lh + wh); leave out any face that is open. (No — derive it)
Increasing / decreasing condition
Increasing: f′(x) > 0. Decreasing: f′(x) < 0. (No — learn it)
Sign of dy/dx either side of a maximum
+ then −: up, flat, down.
Sign of dy/dx either side of a stationary point of inflection
The same on both sides: + 0 + or − 0 −.
A repeated root of y = f(x), e.g. y = x(x − 3)² at x = 3
The curve touches the x-axis there, so the point is also a stationary point.
Positive cubic with two turning points: which comes first?
The maximum is to the left of the minimum.
Where the graph of f′(x) is below the x-axis
f is decreasing there; where it crosses the axis, f has a stationary point.
Graph of y = x + 9/x
Two branches with asymptotes x = 0 and y = x; maximum (−3, −6) lower than minimum (3, 6).
Why can a maximum be lower than a minimum?
Maxima and minima are local: each is the highest or lowest point only near itself.
What does d²y/dx² = 0 at a stationary point tell you?
Nothing conclusive. Check the sign of dy/dx either side (y = x⁴ has d²y/dx² = 0 at a minimum).
Why must an optimisation formula have only one variable?
You can only differentiate with respect to one variable; use the constraint to eliminate the other.
How do you show f(x) is increasing for all x?
Show f′(x) > 0 for every x, e.g. complete the square: a(x + p)² + q with a > 0, q > 0.
Why reject some solutions in a practical problem?
Lengths must be positive and dimensions must be possible (e.g. a cut-out cannot exceed half the card's width).
Stationary points of y = x³ − 3x²
dy/dx = 3x(x − 2) = 0: (0, 0) and (2, −4).
Nature of the stationary points of y = x³ − 3x²
d²y/dx² = 6x − 6: −6 at x = 0 (maximum), 6 at x = 2 (minimum).
Minimum point of y = x + 4/x, x > 0
1 − 4/x² = 0 ⇒ x = 2; y = 4. d²y/dx² = 8/x³ = 1 > 0, so (2, 4) is a minimum.
Where is f(x) = x³ − 27x decreasing?
f′(x) = 3x² − 27 < 0 ⇒ −3 < x < 3.
Show f(x) = x³ + 6x² + 15x is increasing
f′(x) = 3x² + 12x + 15 = 3(x + 2)² + 3 ≥ 3 > 0 for all x.
A rectangle has perimeter 100 m. Largest area?
A = x(50 − x); dA/dx = 50 − 2x = 0 ⇒ x = 25; A = 625 m² (a square).
Open box, square base x, volume 500 cm³: least outer surface area
S = x² + 2000/x; 2x − 2000/x² = 0 ⇒ x = 10; S = 300 cm².
Substituting the stationary x-value into dy/dx to get y
That always gives 0. Substitute into the equation of the curve.
Losing the negative root of x² = 9
Both x = 3 and x = −3 give stationary points unless the domain rules one out.
Writing “minimum” with no reason
State the value or sign: “d²y/dx² = 2/3 > 0, so minimum”.
Stopping once x is found in an optimisation question
Re-read the question: it usually wants the maximum area, the volume, the dimensions or the cost.
“Show that” formula parts
Write the constraint, rearrange, substitute and simplify to exactly the given form, every step visible.
Rounding r before working out the minimum cost
Use the unrounded value in the final calculation, then round the answer.

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