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Stationary points, maxima and minima
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At a stationary point the gradient is zero. Calculus finds the highest and lowest points of a curve, and the best answer to practical problems.
Finding stationary points
Solve dydx = 0, then substitute each x into y.
y = x3 − 3x2 − 9x + 2
dydx = 3x2 − 6x − 9 = 3(x − 3)(x + 1) = 0, so x = −1 or x = 3.
y = x3 − 3x2 − 9x + 2
dydx = 3x2 − 6x − 9 = 3(x − 3)(x + 1) = 0, so x = −1 or x = 3.
Maximum or minimum?
Find d2ydx2, the second derivative.
d2ydx2 < 0: maximum
d2ydx2 > 0: minimum
Here d2ydx2 = 6x − 6: at x = −1 it is −12 (maximum), at x = 3 it is 12 (minimum).
You must give this justification in an exam.
d2ydx2 < 0: maximum
d2ydx2 > 0: minimum
Here d2ydx2 = 6x − 6: at x = −1 it is −12 (maximum), at x = 3 it is 12 (minimum).
You must give this justification in an exam.
Practical problems
Write the quantity to optimise in terms of one variable, differentiate, set to zero, and justify.
60 m of fence makes three sides of a rectangle against a wall: A = x(60 − 2x)
dAdx = 60 − 4x = 0 gives x = 15, and d2Adx2 = −4 < 0, so A = 450 m2 is a maximum.
60 m of fence makes three sides of a rectangle against a wall: A = x(60 − 2x)
dAdx = 60 − 4x = 0 gives x = 15, and d2Adx2 = −4 < 0, so A = 450 m2 is a maximum.
Find the minimum value of y = x + 9x for x > 0.
- dydx = 1 − 9x2 = 0, so x2 = 9 and x = 3
- d2ydx2 = 18x3 > 0 at x = 3, so a minimum
- y = 3 + 3
Answer: 6
Stationary points: dydx = 0. Second derivative negative means maximum, positive means minimum. Always justify the nature.
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