Surface area of a cuboid · 5 marks

Find a cuboid with edges whose lengths are integers (in cm), given that it has a total surface area of exactly 100 cm².

(i) Is there more than one such cuboid?
(ii) Can you find them all?
(iii) Show that you have found them all.

Answer: (i) Yes: for example 1 cm × 2 cm × 16 cm and 2 cm × 4 cm × 7 cm. (ii) These two are the only ones. (iii) With l ≤ w ≤ h, lw + wh + hl = 50 gives (w + l)(h + l) = 50 + l² and l ≤ 4; checking l = 1, 2, 3, 4 gives only these two.

Step-by-step solution

Idea: TSA = 100 means lw + wh + hl = 50. Order the edges l ≤ w ≤ h so that each cuboid is counted once. Adding l² to both sides makes the left side factorise: (w + l)(h + l) = 50 + l².

(i) Is there more than one such cuboid?

1 × 2 × 16: TSA = 2(1 × 2 + 2 × 16 + 16 × 1) = 2(2 + 32 + 16) = 100 cm² ✓.½ mark
2 × 4 × 7: TSA = 2(8 + 28 + 14) = 2 × 50 = 100 cm² ✓. So yes, there is more than one.½ mark

Yes: 1 cm × 2 cm × 16 cm and 2 cm × 4 cm × 7 cm both have TSA 100 cm².

(ii) Can you find them all?

All such cuboids (edges in cm, listed smallest first): 1 × 2 × 16 and 2 × 4 × 7. There are no others; part (iii) proves this.1 mark

Exactly two: 1 × 2 × 16 and 2 × 4 × 7.

(iii) Show that you have found them all.

Let the edges be whole numbers with l ≤ w ≤ h. TSA = 100 ⇒ lw + wh + hl = 50.½ mark
l is at most 4: each of lw, wh, hl is at least l², so 3l² ≤ 50, l² ≤ 16, l ≤ 4.½ mark
A useful form: add l² to both sides: wh + lw + lh + l² = 50 + l², i.e. (w + l)(h + l) = 50 + l². Look for factor pairs with w + l ≤ h + l and w ≥ l.½ mark
l = 1: (w + 1)(h + 1) = 51 = 3 × 17 (51 × 1 would need w = 0) ⇒ w = 2, h = 16.
l = 2: (w + 2)(h + 2) = 54, with w + 2 ≥ 4: 6 × 9 ⇒ w = 4, h = 7. (54 = 2 × 27 or 3 × 18 gives w < 2.)1 mark
l = 3: (w + 3)(h + 3) = 59, a prime, so no solution.
l = 4: (w + 4)(h + 4) = 66 with both factors at least 8; 66 = 6 × 11 = 3 × 22 = 2 × 33 = 1 × 66 has no such pair. So no solution.
Hence only 1 × 2 × 16 and 2 × 4 × 7.½ mark

Every case l = 1, 2, 3, 4 was checked; only 1 × 2 × 16 and 2 × 4 × 7 work.

(i) Yes. (ii) The cuboids are 1 cm × 2 cm × 16 cm and 2 cm × 4 cm × 7 cm. (iii) With l ≤ w ≤ h, (w + l)(h + l) = 50 + l² and l ≤ 4; checking l = 1, 2, 3, 4 gives only these two.

Check: A computer search over all whole-number edges up to 50 finds exactly these two cuboids ✓.

Answer to write in the exam

(i)

1 × 2 × 16: TSA = 2(2 + 32 + 16) = 100 cm²

2 × 4 × 7: TSA = 2(8 + 28 + 14) = 100 cm²

∴ Yes, more than one cuboid

(ii)

All cuboids: 1 cm × 2 cm × 16 cm and 2 cm × 4 cm × 7 cm

(iii)

Let l ≤ w ≤ h; lw + wh + hl = 50

3l² ≤ 50 ⇒ l ≤ 4

(w + l)(h + l) = 50 + l²

l = 1: (w + 1)(h + 1) = 51 = 3 × 17 ⇒ 1 × 2 × 16

l = 2: (w + 2)(h + 2) = 54 = 6 × 9 ⇒ 2 × 4 × 7

l = 3: 59 is prime ⇒ none; l = 4: 66 has no factor pair with both factors ≥ 8 ⇒ none

∴ Only 1 × 2 × 16 and 2 × 4 × 7

Common mistakes that cost marks

Forgetting that TSA = 2(lw + wh + hl), so the three face areas must add to 50, not 100.
Giving a few examples and calling it a proof. You must show that no other case can work, by limiting l and checking every case.
Counting 1 × 2 × 16 and 16 × 2 × 1 as different cuboids. They are the same cuboid turned around.

How this can come in the exam

MCQ (1 mark)

Which cuboid has a total surface area of 48 cm²?

2 cm × 2 cm × 5 cm
1 cm × 2 cm × 8 cm
1 cm × 3 cm × 6 cm
2 cm × 3 cm × 4 cm

Show answer

(A) 2 cm × 2 cm × 5 cm
2(4 + 10 + 10) = 48 cm². The others give 2(2 + 16 + 8) = 52, 2(3 + 18 + 6) = 54 and 2(6 + 12 + 8) = 52.

Try one yourself

Find all cuboids with whole-number edges (in cm) whose total surface area is 22 cm².