2021 ONE(e)

Excellence
Question
A cone has a height of 3 m and a radius of 1.5 m.

A cylinder is inscribed in the cone, as shown in the diagram below.
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The base of the cylinder has the same centre as the base of the cone.

Prove that the maximum volume of the cylinder is π\pi m3^3.

*You must use calculus and show any derivatives that you need to find when solving this problem.*
Official Answer
V=πr2hV=\pi r^2h
=πr2(32r)=\pi r^2(3-2r)
=3πr22πr3=3\pi r^2-2\pi r^3
dVdr=6πr6πr2\dfrac{dV}{dr}=6\pi r-6\pi r^2

At maximum, dVdr=0\dfrac{dV}{dr}=0

6πr(1r)=06\pi r(1-r)=0
r=0r=0 (no) r=1\therefore r=1
V=π12(32×1)=πV=\pi 1^2(3-2\times 1)=\pi
d2Vdr2=6π12πr\dfrac{d^2V}{dr^2}=6\pi-12\pi r

When r=1r=1, d2Vdr2=6π<0\dfrac{d^2V}{dr^2}=-6\pi<0

Therefore V=πV=\pi
is maximum volume.
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Grading Criteria

Achievement (u)

  • Correct expression for dVdr\dfrac{dV}{dr}.

Merit (r)

  • Correct expression for dVdr\dfrac{dV}{dr} and finds r=1r=1.

Excellence T1

  • T1: Correct expression for dVdr\dfrac{dV}{dr} and shows that V=πV=\pi but does not prove it is the maximum volume with either the first or second derivative test.

Excellence T2

  • T2: Correct expression for dVdr\dfrac{dV}{dr} and correct proof.
Video Explanation
NCEA Level 3 Calculus Differentiation 2021 NZQA Exam - Worked Answers by infinityplusone(starts at 21:57)
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