NCEA Calculus 2019 THREE(c - Solution & Explanation

Question

A rectangle has one vertex at

(0,0)

, and the opposite vertex on the curve

y = 4 - \sqrt{x}

, where

0 < x < 16

, 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

A(x)=x\left(4-\sqrt{x}\right)

=4x-x^{\frac{3}{2}}

A'(x)=4-\frac{3}{2}x^{\frac{1}{2}}

Maximum area when

A'(x)=0

\frac{3}{2}\sqrt{x}=4

\sqrt{x}=\frac{8}{3}

x=\frac{64}{9}

Area

=\frac{64}{9}\left(4-\frac{8}{3}\right)

=\frac{64}{9}\times\frac{4}{3}

=\frac{256}{27}\ \left(=9\ \frac{13}{27}\right)

Accept

9.48

Grading Criteria

Achievement (u)

Correct expression for $A'(x)$ .

Merit (r)

Correct solution with correct derivative.

Excellence T1

-

Excellence T2

-

Video Explanation

NCEA Level 3 Calculus Differentiation 2019 NZQA Exam - Worked Answers by infinityplusone(starts at 50:19)

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2019 THREE(c)

Achievement (u)

Merit (r)

Excellence T1

Excellence T2