Optimisation & Rates of Change (Grade 12) — Worksheet & Memo

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Grade 12 · Calculus

Optimisation & Rates of Change (Grade 12)

MARKING GUIDELINE

Marks

Duration

50 minutes

Questions

Name:

Class:

Date:

Mark

/ 31

Instructions and Information

Answer ALL the questions in this question paper.
Answer QUESTION 1 by circling the letter (A–D) in the answer grid at the end of that section.
Show ALL calculations clearly.
Show all units where applicable.
Number the answers correctly according to the numbering system used in this question paper.
A non-programmable calculator may be used, unless stated otherwise.
Write neatly and legibly.

Question 1

[10 MARKS]

Four options are given as possible answers to the following questions. Choose the correct answer and circle the letter (A–D) in the grid at the end of this section. If you want to change your choice, put a cross through the wrong letter and circle your new choice.

1.1
To MAXIMISE or MINIMISE a quantity, you …
(1)
A)set the expression itself equal to 0
B)differentiate the expression and set the derivative equal to 0✓
C)substitute the largest allowed value
D)set the second derivative equal to 0
Answer: B — A maximum or minimum occurs where the rate of change is zero.
- A — that finds where the quantity is zero, not where it is largest
- C — the optimum need not lie at an endpoint
- D — setting f″ to zero finds the point of INFLECTION
1.2
A rectangle has a perimeter of 40 m. Its greatest possible AREA is …
(1)
A)400 m²
B)100 m²✓
C)40 m²
D)200 m²
Answer: B — A = l(20 − l), and dA/dl = 20 − 2l = 0 gives l = 10, so A = 100 m².
- A — squared half the perimeter instead of optimising
- C — used the perimeter as though it were the area
- D — halved the perimeter and multiplied by 10 without optimising
1.3
A closed box has a square base of side x and a volume of 500 cm³. Its height is …
(1)
A)500/x
B)500x²
C)x²/500
D)500/x²✓
Answer: D — V = x²h = 500, so h = 500 ÷ x².
- A — the base is a SQUARE, so its area is x², not x
- B — multiplied instead of dividing
- C — inverted the expression
1.4
If s(t) = 5t² + 2t metres, the VELOCITY at t = 3 seconds is …
(1)
A)30 m per second
B)51 m per second
C)10 m per second
D)32 m per second✓
Answer: D — v = s′(t) = 10t + 2, so v(3) = 32.
- A — forgot to add the 2
- B — substituted into s(t) instead of s′(t): 45 + 6 = 51
- C — gave the coefficient without substituting t = 3
1.5
ACCELERATION is the derivative of …
(1)
A)velocity with respect to time✓
B)velocity with respect to distance
C)time with respect to velocity
D)distance with respect to time
Answer: A — a = dv/dt, which is the second derivative of displacement.
- B — acceleration is measured against TIME
- C — that inverts the relationship
- D — that is VELOCITY
1.6
The sum of two positive numbers is 30. Their product is largest when the numbers are …
(1)
A)15 and 15✓
B)1 and 29
C)0 and 30
D)10 and 20
Answer: A — P = x(30 − x), and dP/dx = 30 − 2x = 0 gives x = 15, so the product is 225.
- B — 1 × 29 = 29, far below 225
- C — 0 × 30 = 0, the smallest product possible
- D — 10 × 20 = 200, still below 225
1.7
A cylinder of volume 1 000 cm³ has radius r. Its height is …
(1)
A)(1 000)/πr
B)1 000πr²
C)(1 000)/πr²✓
D)(1 000)/r²
Answer: C — V = πr²h = 1 000, so h = 1 000 ÷ (πr²).
- A — the base area is πr², not πr
- B — multiplied instead of dividing
- D — left out the π
1.8
At a MAXIMUM turning point, f″(x) is …
(1)
A)undefined
B)positive
C)negative✓
D)zero
Answer: C — A maximum is concave down, so the second derivative is negative there.
- A — the second derivative of a polynomial exists everywhere
- B — a positive second derivative means a MINIMUM
- D — at a maximum the curve is concave DOWN, so f″ is negative; f″ = 0 is where a cubic changes concavity
1.9
A stone's height is h = 30t − 5t² metres after t seconds. Its greatest height is …
(1)
A)30 m
B)3 m
C)90 m
D)45 m✓
Answer: D — h′(t) = 30 − 10t = 0 gives t = 3, and h(3) = 90 − 45 = 45 m.
- A — gave the coefficient 30 instead of a height
- B — that is the TIME at the maximum, not the height
- C — used 30t alone and forgot to subtract the 5t²
1.10
In an optimisation problem, the CONSTRAINT is used to …
(1)
A)write the quantity in terms of a single variable✓
B)avoid using calculus at all
C)find the derivative directly
D)check the answer at the very end only
Answer: A — In Grade 12 optimisation the quantity is first written in terms of ONE variable, and the constraint is what removes the other.
- B — calculus is still what finds the optimum
- C — the derivative can only be taken once one variable is left
- D — the constraint is needed BEFORE differentiating, not only afterwards

Answer grid — marking guideline

1.1	A	B	C	D
1.2	A	B	C	D
1.3	A	B	C	D
1.4	A	B	C	D
1.5	A	B	C	D
1.6	A	B	C	D
1.7	A	B	C	D
1.8	A	B	C	D
1.9	A	B	C	D
1.10	A	B	C	D

Question 2

[13 MARKS]

Given f(x) = x². Determine the MINIMUM distance between the point (10 ; 2) and a point on f.

2.1
Write down an expression for the square of the distance d between (10 ; 2) and the point (x ; x²) on f.
(3)
d² = (x − 10)² + (y − 2)² (1)
y = x² on f (1)
d² = (x − 10)² + (x² − 2)² (1)
2.2
Explain why minimising d² gives the same answer as minimising d.
(2)
d is never negative (1), and squaring preserves the order of non-negative numbers, so both are smallest at the same x (1)
2.3
Determine the value of x for which the distance is a minimum.
(5)
d² = x⁴ − 3x² − 20x + 104 (1)
(d(d²))/dx = 4x³ − 6x − 20 (2)
Set it equal to 0 and test factors: x = 2 gives 32 − 12 − 20 = 0 (1)
x = 2 (1)
2.4
Hence calculate the minimum distance, in simplest surd form.
(3)
At x = 2 the point on f is (2 ; 4) (1)
d² = (−8)² + 2² = 68 (1)
d = 2√17 units (1)

Question 3

[8 MARKS]

A container is being filled with water. After t seconds the volume, in litres, is given by V(t) = 3t² + 2t.

3.1
Determine the AVERAGE rate at which the volume increases over the first 5 seconds.
(3)
V(0) = 0 and V(5) = 85 (1)
(85 − 0)/(5 − 0) (1) = 17 litres per second (1)
3.2
Determine the rate at which the volume is increasing AT t = 5 seconds.
(3)
V′(t) = 6t + 2 (2)
V′(5) = 32 litres per second (1)
3.3
Explain why the two answers differ.
(2)
The first is the average over the whole interval (1); the second is the instantaneous rate at one moment, and the flow is speeding up (1)

TOTAL: 31 marks

This question paper consists of 3 questions.