Question 3(e) Merit to Excellence

Reading differentiability, stationary points and a limit straight off a drawn graph, then a log quotient, a related-rates sphere, a parametric gradient, and a rectangle inscribed in a curve.

The diagram shows part of the symmetrical graph y² = 16x − x². A rectangle ABCD is drawn inside the curve with vertices B and C on the curve.

Find the length AD so that the rectangle has its maximum area. You can assume that the value found is a maximum.

The 2025 paper, page 12 ↗ Marking schedule, page 6 ↗

Work it out

Try it on paper first. Stuck? Take one hint at a time.

1. 1 Put A at (x, 0). Where are D, B and C?

   The curve meets the x-axis at 0 and 16, and is symmetrical about x = 8. So D is at (16 − x, 0), and B and C are at height y = √(16x − x²).

2. 2 Write the area as a function of x alone.

   AD = 16 − 2x and AB = (16x − x²)^1/2
   Area = (16 − 2x)(16x − x²)^1/2

3. 3 Differentiate with the product rule. Where does the chain rule come in?

   On the square root:
   dA/dx = −2(16x − x²)^1/2 + (16 − 2x) · (1/2)(16x − x²)^−1/2(16 − 2x)
   = −2(16x − x²)^1/2 + ((8 − x)(16 − 2x))/(16x − x²)^1/2

4. 4 Set it to zero, then clear the square root.

   Multiply through by (16x − x²)^1/2:
   2(16x − x²) = (8 − x)(16 − 2x) = 2(8 − x)²
   16x − x² = 64 − 16x + x²
   2x² − 32x + 64 = 0, so x² − 16x + 32 = 0

5. 5 Solve it. Which root fits the diagram?

   x = 8 ± 4√2, that is 2.343 or 13.657. A is left of the middle, so x = 8 − 4√2 ≈ 2.343. (13.657 just swaps A and D.)

6. 6 The question asks for AD, not x.

   AD = 16 − 2(8 − 4√2) = 8√2 ≈ 11.31

What the marker wanted

• Achieved dA/dx from an area model, even with one minor error in it.
• Merit A correct dA/dx.
• Excellence AD ≈ 11.314, with the calculus set out in full (E8). Getting as far as x, or a correct solution with one minor error, is E7.

In our words, from NZQA’s 2025 marking schedule.

Common mistake

How did it go?

Got it moves it further out. Review brings it back tomorrow.