IB Mathematics SL 5.7 Optimization problems in context AI SL Paper 1- Exam Style Questions- New Syllabus

(a)
At maximum volume, \( \frac{dV}{dw} = 0 \).
\( 690w – 36w^2 = 0 \)
\( w(690 – 36w) = 0 \)
\( w = 0 \) or \( w = \frac{690}{36} = \frac{115}{6} \approx 19.1666… \).
\( w = 0 \) gives minimum volume (zero), so maximum occurs at \( w = \frac{115}{6} \).
✅ Answer: \(\boxed{19.2 \text{ cm}}\) (accept \(19.2\), \(19.17\), or \(\frac{115}{6}\)).

(b)
Integrate \( \frac{dV}{dw} \) to find \( V \):
\( V = \int (690w – 36w^2) \, dw = 345w^2 – 12w^3 + c \).
When \( w = 0 \), volume is zero, so \( c = 0 \).
Thus \( V = 345w^2 – 12w^3 \).
Substitute \( w = \frac{115}{6} \):
\( w^2 = \frac{13225}{36} \), \( w^3 = \frac{1520875}{216} \).
\( V = 345 \times \frac{13225}{36} – 12 \times \frac{1520875}{216} \)
Simplify: \( V = \frac{4562625}{36} – \frac{18250500}{216} \).
Common denominator 216: \( V = \frac{27375750}{216} – \frac{18250500}{216} = \frac{9125250}{216} \approx 42246.527… \).
To three significant figures: \( 42200 \) but careful rounding: \( 42246.5 \to 42200\)? Wait, 42246.5 to 3 s.f. is \(42200\) (since the fourth digit is 4).
✅ Answer: Maximum volume is \( \boxed{42200 \text{ cm}^3} \) (or \(4.22 \times 10^4\)).

(c)
Given length \( l = 3w \), and height \( h \) unknown.
Volume formula: \( V = l \times w \times h = 3w^2 h \).
From part (b), at maximum volume: \( V = 42200 \), \( w = 19.1666… \).
Thus \( h = \frac{V}{3w^2} = \frac{42200}{3 \times (19.1666…)^2} \approx 38.3 \text{ cm} \).
Now \( M = w + l + h = w + 3w + h = 4w + h \).
\( M = 4 \times 19.1666… + 38.3 \approx 76.666… + 38.3 = 114.966… \approx 115 \).
Alternatively, from the integrated volume expression \( V = 345w^2 – 12w^3 \), and \( h = \frac{V}{3w^2} = 115 – 4w \).
Then \( M = w + 3w + (115 – 4w) = 115 \).
✅ Answer: \(\boxed{115 \text{ cm}}\).