2018 THREE(e)

Excellence
Question
Loading diagram...
The above shape is made from wire. It has both vertical and horizontal lines of symmetry.

The ends of the shape are at the vertices of a square with a side length of 10 cm, as shown in
the diagram above.

The length of the piece of wire through the centre of the shape is xx cm.

Find the value(s) of xx that enables the shape to be made with the minimum length of wire.
You do not need to prove that the length is a minimum.

*You must use calculus and show any derivatives that you need to find when solving this
problem.*
Official Answer
w2=52+(5x2)2w^2=5^2+\left(5-\dfrac{x}{2}\right)^2
w2=25+255x+0.25x2w^2=25+25-5x+0.25x^2
w2=0.25x25x+50w^2=0.25x^2-5x+50
w=0.25x25x+50w=\sqrt{0.25x^2-5x+50}
Length =x+4w=x+4w
=x+40.25x25x+50=x+4\sqrt{0.25x^2-5x+50}
dLdx=1+2(0.25x25x+50)12×(0.5x5)\dfrac{dL}{dx}=1+2\left(0.25x^2-5x+50\right)^{-\frac{1}{2}}\times(0.5x-5)
dLdx=1+x100.25x25x+50\dfrac{dL}{dx}=1+\dfrac{x-10}{\sqrt{0.25x^2-5x+50}}
For max/min dLdx=0\dfrac{dL}{dx}=0
x100.25x25x+50=1\dfrac{x-10}{\sqrt{0.25x^2-5x+50}}=-1
x10=10.25x25x+50x-10=-1\sqrt{0.25x^2-5x+50}
(x10)2=0.25x25x+50(x-10)^2=0.25x^2-5x+50
x220x+100=0.25x25x+50x^2-20x+100=0.25x^2-5x+50
0.75x215x+50=00.75x^2-15x+50=0
x=15.77x=15.77 not applicable
x=4.23 cmx=4.23\text{ cm}
Grading Criteria

Achievement (u)

-

Merit (r)

  • Correct expression for dLdx\dfrac{dL}{dx}

Excellence T1

  • Correct solution with correct derivative.

Excellence T2

-
Video Explanation
NCEA Level 3 Calculus Differentiation 2018 NZQA Exam - Worked Answers by infinityplusone(starts at 51:40)
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