A closed cylindrical can with base radius r cm and height h cm is constructed from a thin sheet of metal to hold 50π cm³ of liquid.
(a) Show that the total area, A cm², of material needed to make this can is given by A = 2πr² + 100π/r.
(b) Find the values of r and h if the area of material used is to be the least.
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Answer:
(a) Volume of the cylindrical can = πr²h
πr²h = 50π
h = 50/r²
Total surface area of cylindrical can,
A = 2πr² + 2πrh
= 2πr² + πr(50/r²)
= 2πr² + 100π/r
(shown)
(b) A = 2πr² + 100π/r
dA/dr = 4πr - 100π/r²
For maximum or minimum values,
dA/dr = 0
4πr - 100π/r² = 0
4πr = 100π/r²
r³ = 25
When
= 12π
>0
==> A is a minimum when
Substituting
