All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
Volume, V = \pi r^2 h + \frac{2}{3} \pi r^2 = k

\Rightarrow h = \frac{3k - 2\pi r^3}{3 \pi r^2}

Surface Area, A =  2 \pi r^2 + 2 \pi rh + \pi r^2

= 3 \pi r^2 + 2 \pi r \frac{3k - 2\pi r^3}{3 \pi r^2}

= \frac{5 \pi r^2}{3} + \frac{2k}{r}

\frac{dA}{dr} = \frac{10 \pi r}{3} - \frac{2k}{r^2}

Set \frac{dA}{dr} = 0

\Rightarrow r = (\frac{3k}{5 \pi})^{\frac{1}{3}}

\frac{d^2A}{dr^2} = \frac{10\pi}{3} + \frac{4k}{r^3} = 10 \pi > 0 when r = (\frac{3k}{5 \pi})^{\frac{1}{3}}.

Thus A is minimum when r = (\frac{3k}{5 \pi})^{\frac{1}{3}}

\Rightarrow h = r = (\frac{3k}{5 \pi})^{\frac{1}{3}}

(ii)
Surface Area, A = \frac{5 \pi r^2}{3} + \frac{400}{r} = 180

\Rightarrow 5\pi r^2 + 1200 - 540 r = 0

Using the Graphing Calculator, we find that r = 3.037205 \mathrm{~or~} 3.72153

h = 4.88 \mathrm{~or~} 2.12

Since r \textless h, \mathrm{~then~}  r = 3.04, h = 4.88

KS Comments:

Notice that the value of r and h is the same. Which actually makes some intuitive sense if you attempt to visualise. Students should simplify to the simplest form, using appropriate indices formula.
For (ii), it can be solved using GC entirely. If students can, they should draw the curve out.

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