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Monday, June 20, 2011

A Max-min Quiz


Given V  = Vcylinder + Vhemisphere = 2880π 
a) Make h the subject of the equation:
2880π =  πr2h + (2/3) πr3  
Divide both sides by π
2880 = r2h + (2/3) r3  
h = {2880 – (2/3) r3}/ r2 
h = 2880/r2  – ()r = 2880r–2 – ()r
b) Find the surface area as a function of r only.
Surface area, S = Acircle + Arectangle + Ahemisphere
S = πr2 + 2πrh + 2πr2
S = 3πr2 + 2πrh
S = 3πr2 + 2πr{2880r–2 – ()r}
S = 3πr2 + 5760πr–1 – (4π/3) r2
S = (5π/3) r2 + 5760πr–1                                                                           as required
c) Find the derivative of the surface area, the stationary point of the surface area function and the surfasce area at the stationary point.
dS/dr = 2(5π/3) r1 + (-1)5760πr–2 
dS/dr = (10 π /3) r – 5760πr–2 
dS/dr = (10 π /3) r – 5760πr–2 
find a common denominator: dS/dr = (10 π r3/3r2)– (17280π/3r2)
For a stationary point dS/dr = 0
For a fraction to equal 0, the numerator equals 0.
Therefore 0 = 10 π r3 – 17280π
17280 π = 10 π r3
r3  = 1728
r = 12
When r = 12, S = (5π/3) (12)2 + 5760π/12 
S = 240 π + 480 π
S = 760 π   (~ 2260 but leave in terms of π)
A sketch:


From the graph the stationary point appears to be a minimum. To confirm this, find the second derivative:
d2S/dr2 = 10 /3 + 11520 /r3
d2S/dr2 = a  positive term + another positive term
Hence the curve is concave up.  It is confirmed that the stationary point is a minimum.

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