IB · Mathematics: Analysis & Approaches SL & HL · Calculus

Stationary points & optimisation: IB Mathematics: Analysis & Approaches questions with answers

Free Stationary points & optimisation questions for IB Mathematics: Analysis & Approaches (SL & HL), part of Calculus. Each question has the answer and the marking points an examiner would look for.

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Stationary points & optimisation practice questions

Question 1Structured · 6 marksA* challenge

A window is in the shape of a rectangle of width x m with a semicircle on top. The perimeter of the window is 6 m.

(a)

Show that the area of the window is A = 3x − x22 − πx28.

[3]
(b)

Find the width that maximizes the area, and the maximum area.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)shown
x + 2h + πx2 = 6 A1
A = xh + πx28 with h substituted M1
simplifying AG A1
3
(b)x = 1.68 m, A = 2.52 m2
attempt to find the maximum, e.g. A′(x) = 0 or graph (M1)
x = 1.68 A1
A = 2.52 m2 A1
3
Question 2Structured · 7 marks

Let f(x) = x3 + ax2 + bx − 4. The graph of f has stationary points at x = 0 and x = 6.

(a)

Find the value of a and the value of b.

[4]
(b)

Find the coordinates of the local maximum point, justifying that it is a maximum.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)a = −9, b = 0
f′(x) = 3x2 + 2ax + b A1
f′(0) = 0 and f′(6) = 0, or f′(x) = 3(x)(x − 6) M1
a = −9 A1
b = 0 A1
4
(b)(0, −4)
f″(x) = 6x − 18 (A1)
f″(0) = −18 < 0, so a maximum R1
(0, −4) A1
3
Question 3Structured · 14 marks

The concentration C mg l−1 of a drug in the blood t hours after it is taken is modelled by C(t) = 50te−0.4t, t ≥ 0.

(a)

Find C′(t).

[3]
(b)

Find the time at which the concentration is greatest and the maximum concentration.

[4]
(c)

The drug is effective while the concentration is above 13.8 mg l−1. Find the length of time for which the drug is effective.

[4]
(d)

Find the rate of change of the concentration when t = 10, and interpret its sign.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)50e−0.4t(1 − 0.4t)
product rule M1
50e−0.4t − 20te−0.4t A1A1
3
(b)t = 2.50 h, C = 46.0 mg l−1
C′(t) = 0 M1
t = 10.4 = 2.50 A1
substituting (M1)
C = 46.0 A1
4
(c)8.28 h
C(t) = 13.8 (M1)
t = 0.313 A1
t = 8.60 A1
8.28 h A1
4
(d)−2.75 mg l−1 h−1; decreasing
C′(10) (M1)
−2.75 A1
negative, so the concentration is decreasing R1
3
Question 4Structured · 6 marks

An open box is made from a square sheet of card with sides 18 cm by cutting a square of side x cm from each corner and folding up the sides.

(a)

Show that the volume of the box is V = x(18 − 2x)2 cm3.

[1]
(b)

Find the value of x that maximizes the volume, and the maximum volume.

[5]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)shown
base side 18 − 2x, height x AG A1
1
(b)x = 3, V = 432 cm3
dVdx = (18 − 2x)2 − 4x(18 − 2x) M1A1
(18 − 2x)(18 − 6x) = 0 (A1)
x = 3 (x = 9 gives zero volume) A1
V = 432 cm3 A1
5
Question 5Structured · 14 marksA* challenge

A farmer has 300 m of fencing to enclose a rectangular field next to a straight river. No fencing is needed along the river. The side parallel to the river has length x m and the other two sides have length y m.

(a)

Show that the area of the field is A = x(300 − x)2.

[2]
(b)

Find the maximum area, justifying that it is a maximum.

[4]
(c)

A second field must have area 3200 m2. The fence parallel to the river costs $12 per metre and the other two sides cost $10 per metre. Show that the cost is C(x) = 12x + 64000x.

[3]
(d)

Find the dimensions that minimize the cost, and the minimum cost.

[5]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)shown
x + 2y = 300 A1
A = xy with y substituted AG A1
2
(b)11250 m2
A′(x) = 150 − x A1
A′(x) = 0 ⇒ x = 150 M1
A″(x) = −1 < 0 R1
11250 m2 A1
4
(c)shown
y = 3200x A1
C = 12x + 2 × 10y M1
substituting AG A1
3
(d)x = 73.0 m, y = 43.8 m, $1753
C′(x) = 0 or graph (M1)
x = 73.0 A1
y = 43.8 A1
substituting (M1)
$1753 A1
5
Question 6Structured · 7 marksA* challenge

A closed cylindrical can has radius r cm and height h cm. Its volume is 250π cm3.

(a)

Show that the surface area is S = 2πr2 + 500πr.

[2]
(b)

Find the value of r that minimizes S, and the minimum surface area, justifying that it is a minimum.

[5]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)shown
h = 250r2 A1
S = 2πr2 + 2πrh, substituted AG A1
2
(b)r = 5, S = 150π
dSdr = 4πr − 500πr2 A1
setting equal to 0 M1
r3 = 125 ⇒ r = 5 A1
d2Sdr2 > 0, so a minimum R1
S = 150π cm2 A1
5

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Original IB-style practice questions written for mentorriyaz.in. These are not IB past paper questions, and this site is not affiliated with or endorsed by the International Baccalaureate Organization. Updated 2026-11-04. Paper generator · mentorriyaz.in