NCEA Calculus 2023 ONE(e - Solution & Explanation

Question

The graph of

y = x(x - 2m)^2

, where

m > 0

, is shown.

Loading diagram...

The total shaded area between the curve and the

x

-axis
from

x = 0

to

x = 2m

is given by

A = \dfrac{4m^4}{3}

.

A right-angled triangle is now constructed with one
vertex at

(0,0)

and another on the curve

y = x(x - 2m)^2

,
as shown below.

Loading diagram...

Show that the maximum area of such a triangle is

\dfrac{3}{8}

of the total shaded area.

*You must use calculus and show any derivatives that you need to find when solving this problem.*
*You do not have to prove that the area you have found is a maximum.*

Official Answer

Area of triangle

=\dfrac{1}{2}xy

A=\dfrac{1}{2}x\big(x(x-2m)^2\big)

=\dfrac{1}{2}x^2(x-2m)^2

\dfrac{dA}{dx}=\dfrac{1}{2}\left(2x(x-2m)^2+x^2\cdot 2(x-2m)\right)

=x(x-2m)^2+x^2(x-2m)

=x(x-2m)\big((x-2m)+x\big)

=x(x-2m)(2x-2m)

=2x(x-2m)(x-m)

OR

A=\dfrac{1}{2}x^2(x-2m)^2

=\dfrac{1}{2}x^4-2mx^3+2m^2x^2

\dfrac{dA}{dx}=2x^3-6mx^2+4m^2x

=2x(x^2-3mx+2m^2)

=2x(x-2m)(x-m)

\dfrac{dA}{dx}=0\Rightarrow 2x(x-2m)(x-m)=0

x=0

or

x=2m

or

x=m

Since

0<x<2m

the area is a maximum when

x=m

Maximum area of triangle:

A(m)=\dfrac{1}{2}m^2(m-2m)^2

=\dfrac{1}{2}m^4

This is

\dfrac{3}{8}

of the total shaded area since

\dfrac{3}{8}\times\dfrac{4m^4}{3}=\dfrac{1}{2}m^4

Grading Criteria

Achievement (u)

Correct derivative.

Merit (r)

Correct derivative.
AND
$x=m$ found.

Excellence T1

T1
Maximum area,
$A=\dfrac{1}{2}m^4$ found
with correct $\dfrac{dA}{dx}$ .

OR
Correct solution but with one minor error.

Excellence T2

T2
Correct solution with correct $\dfrac{dA}{dx}$
showing the calculation of the correct proportion of total shaded area..

Video Explanation

2023 NCEA L3 Calculus Exam Walkthrough by infinityplusone(starts at 18:18)

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2023 ONE(e)

Achievement (u)

Merit (r)

Excellence T1

Excellence T2