NCEA Calculus 2014 ONE(e - Solution & Explanation

Question

What is the maximum volume of a cone if the slant length of the cone is 20 cm?

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You do not need to prove that the volume you have found is a maximum.

*Show any derivatives that you need to find when solving this problem.*

Official Answer

h^2+r^2=400

h=\sqrt{400-r^2}

V=\frac{1}{3}\pi r^2h=\frac{1}{3}\pi r^2\sqrt{400-r^2}

\dfrac{dV}{dr}=\dfrac{2}{3}\pi r\sqrt{400-r^2}+\dfrac{1}{3}\pi r^2\cdot\dfrac{1}{2}(400-r^2)^{-\frac{1}{2}}\cdot-2r

\dfrac{dV}{dr}=\dfrac{\dfrac{2}{3}\pi r(400-r^2)-\dfrac{1}{3}\pi r^3}{\sqrt{400-r^2}}

At maximum volume:

\dfrac{dV}{dr}=0

2(400-r^2)=r^2

3r^2=800

r=16.3\,\text{cm}

V=3225\,\text{cm}^3

Alternative working:

r^2=400-h^2

V=\frac{1}{3}\pi r^2h=\frac{1}{3}\pi(400-h^2)h

=\frac{\pi}{3}(400h-h^3)

\dfrac{dV}{dh}=\frac{\pi}{3}(400-3h^2)

At maximum,

\dfrac{dV}{dh}=0

400-3h^2=0

h^2=\frac{400}{3}

h=\frac{20}{\sqrt{3}}=11.547\,\text{cm}

V=3225\,\text{cm}^3

Grading Criteria

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2014 ONE(e)