IB Mathematics SL 5.7 Optimization problems in context AI HL Paper 2- Exam Style Questions- New Syllabus

a
Split the hexagonal base into 6 equilateral triangles.
Area of one triangle = \( \frac{1}{2} × x^2 × \sin 60^\circ \).
Since \( \sin 60^\circ = \frac{\sqrt{3}}{2} \), area of one triangle = \( \frac{1}{2} × x^2 × \frac{\sqrt{3}}{2} = \frac{\sqrt{3}x^2}{4} \).
Total area = \( 6 × \frac{\sqrt{3}x^2}{4} = \frac{6\sqrt{3}x^2}{4} = \frac{3\sqrt{3}x^2}{2} \).
Explanation:
Use the area formula for an equilateral triangle and multiply by 6.
Result:
\( \frac{3\sqrt{3}x^2}{2} \)

b
Total surface area = 2 × (area of base) + 6 × (lateral area).
Area of base = \( \frac{3\sqrt{3}x^2}{2} \), lateral area per side = \( x × h \), total lateral area = \( 6 × x × h \).
Given \( 1200 = 2 × \frac{3\sqrt{3}x^2}{2} + 6 × x × h \),
\( 1200 = 3\sqrt{3}x^2 + 6xh \),
\( 6xh = 1200 – 3\sqrt{3}x^2 \),
\( h = \frac{1200 – 3\sqrt{3}x^2}{6x} = \frac{400 – \sqrt{3}x^2}{2x} \).
Volume \( V = \text{area of base} × h = \frac{3\sqrt{3}x^2}{2} × \frac{400 – \sqrt{3}x^2}{2x} \),
\( V = \frac{3\sqrt{3}x^2 × (400 – \sqrt{3}x^2)}{4x} = \frac{3\sqrt{3} × 400x – 3\sqrt{3} × \sqrt{3}x^3}{4} \),
\( = \frac{1200\sqrt{3}x – 3 × 3x^3}{4} = \frac{1200\sqrt{3}x – 9x^3}{4} \),
\( V = 300\sqrt{3}x – \frac{9}{4}x^3 \).
Explanation:
Derive h from the surface area equation and substitute into the volume formula.
Result:
\( V = 300\sqrt{3}x – \frac{9}{4}x^3 \)

c

Explanation:
A cubic function with a maximum is sketched for 0 ≤ x ≤ 16.
Result:
Graph as shown

d
\( V = 300\sqrt{3}x – \frac{9}{4}x^3 \),
\( \frac{dV}{dx} = 300\sqrt{3} – \frac{9}{4} × 3x^2 \),
\( \frac{dV}{dx} = 300\sqrt{3} – \frac{27}{4}x^2 \).
Explanation:
Differentiate the volume function with respect to x.
Result:
\( \frac{dV}{dx} = 300\sqrt{3} – \frac{27}{4}x^2 \)

e
Set \( \frac{dV}{dx} = 0 \): \( 300\sqrt{3} – \frac{27}{4}x^2 = 0 \),
\( \frac{27}{4}x^2 = 300\sqrt{3} \),
\( x^2 = \frac{300\sqrt{3} × 4}{27} \),
\( x^2 = \frac{1200\sqrt{3}}{27} \),
\( x = \sqrt{\frac{1200\sqrt{3}}{27}} \approx 8.77382 \), rounded to 8.77.
Explanation:
Solve the derivative equation to find the critical point.
Result:
8.77 (8.77382…)

f
Substitute \( x = 8.77382 \) into \( V = 300\sqrt{3}x – \frac{9}{4}x^3 \),
\( V = 300\sqrt{3} × 8.77382 – \frac{9}{4} × (8.77382)^3 \),
\( 300\sqrt{3} × 8.77382 \approx 4559.34 \),
\( \frac{9}{4} × (8.77382)^3 \approx 1519.996 \),
\( V \approx 4559.34 – 1519.996 \approx 3039.34 \) cm³,
rounded to 3040 cm³.
Explanation:
Evaluate the volume function at the maximum x value.
Result:
3040 cm³ (3039.34…)