Optimisation & Rates of Change (Grade 12)

Building a formula from a described situation, differentiating it to find the maximum or minimum, and interpreting a derivative as a rate of change — including velocity and acceleration.

1The optimisation recipe

Write the quantity to be optimised as a formula.
Use the given constraint to get it in terms of ONE variable.
Differentiate and set the derivative equal to zero.
Solve, reject impossible answers, and answer the question asked.

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Step 2 is where the marks are lost. You cannot differentiate a formula with two variables in it — use the constraint to eliminate one.

Worked ExampleA rectangle has a perimeter of 40 m. Determine the maximum possible area.

1
Let the width be x and the length y. Then 2x + 2y = 40, so y = 20 − x.
The constraint, used to eliminate y.
2
A = xy = x(20 − x) = 20x − x².
Now in one variable.
3
dAdx = 20 − 2x.
4
20 − 2x = 0 → x = 10.
Set the derivative to zero.
5
A = 10(20 − 10) = 100 m².
A square gives the maximum area — as it always does for a fixed perimeter.

2Rates of change

dydx = the rate at which y changes with respect to x

velocity v(t) = s′(t) · acceleration a(t) = v′(t) = s″(t)

Worked ExampleA ball's height is s(t) = 20t − 5t² metres after t seconds. Find its velocity at t = 1 and when it hits the ground.

1
v(t) = s′(t) = 20 − 10t.
Velocity is the derivative of displacement.
2
v(1) = 20 − 10 = 10 m·s⁻¹.
Positive, so it is still rising.
3
It hits the ground when s = 0: 20t − 5t² = 0 → 5t(4 − t) = 0.
4
t = 4 seconds (reject t = 0, the moment it was thrown).
5
v(4) = 20 − 40 = −20 m·s⁻¹.
Negative means it is moving downwards.

The ball reaches its maximum height when v(t) = 0, i.e. 20 − 10t = 0, so t = 2 s and s(2) = 40 − 20 = 20 m.

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A negative velocity does not mean 'slowing down'. It means moving in the opposite direction. The ball at t = 4 is falling fast.

Practice exercises

Work each one out, then click to reveal the answer.

📄 Worksheet ✅ Memo 📝 Exam paper 🗝️ Marking guide

1
A rectangle has a perimeter of 40 m. Determine its maximum possible area.
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A = x(20 − x), A′ = 20 − 2x = 0 → x = 10 → 100 m²
2
The height of a ball is s(t) = 20t − 5t². Determine its velocity after 1 second.
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v(t) = 20 − 10t → v(1) = 10 m·s⁻¹
3
For the same ball, determine when it reaches its maximum height.
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v = 0 → 20 − 10t = 0 → t = 2 s
4
For the same ball, determine its maximum height.
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s(2) = 40 − 20 = 20 m
5
For the same ball, determine its velocity when it strikes the ground.
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s = 0 at t = 4 → v(4) = −20 m·s⁻¹
6
Determine the acceleration of the ball.
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a(t) = v′(t) = −10 m·s⁻² (constant)
7
Why must you use the constraint before differentiating?
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Because you cannot differentiate a formula containing two variables — the constraint eliminates one.
8
What does a negative velocity tell you?
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The object is moving in the opposite direction (downwards, here).