A Max-min Quiz

Given V  = V_cylinder + V_hemisphere = 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  – (⅔)r = 2880r^–2 – (⅔)r

b) Find the surface area as a function of r only.

Surface area, S = A_circle + A_rectangle + A_hemisphere

S = πr² + 2πrh + 2πr²

S = 3πr² + 2πrh

S = 3πr² + 2πr{2880r^–2 – (⅔)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.