2016 THREE(c)

Merit
Question
A rectangle has one vertex at (0,0) and the opposite vertex on the curve y=(x−6)2y = (x - 6)^2, where
0<x<60 < x < 6, as shown on the graph below.
Loading diagram...
Find the maximum possible area of the rectangle.

*You must use calculus and show any derivatives that you need to find when solving this problem.*

*You do not need to prove that the area you have found is a maximum.*
Official Answer
Area=A(x)=x(x−6)2\text{Area}=A(x)=x(x-6)^2
=x3−12x2+36x\qquad\qquad= x^3-12x^2+36x

A′(x)=3x2−24x+36A'(x)=3x^2-24x+36

Max when A′(x)=0\text{Max when }A'(x)=0

3(x2−8x+12)=03(x^2-8x+12)=0

3(x−6)(x−2)=03(x-6)(x-2)=0
Max when x=2\text{Max when }x=2
Max Area=2×16=32\text{Max Area}=2\times 16=32
Grading Criteria

Achievement (u)

  • Correct expression for A′(x)A'(x)

Merit (r)

  • Correct solution for maximum area with correct derivative.

Excellence T1

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Excellence T2

-
Exam Paper (2016)Assessment Schedule (2016)
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