Astandard oil drum is a cylinder with a volume of approximately 218.276 liters (or 0.218276 cubic meters). the cost of producing the cylindrical wall of the oil drum per square meter is the price per square meter of 16-gauge steel sheet metal, plus an additional 10% for the cost of manufacturing. the circular caps of the oil drum are more expensive to produce: they are the price per square meter of 16-gauge steel sheet metal, plus an additional 63.9% for the cost of manufacturing. assuming that the cost of 16-gauge steel sheet metal is $25.00 per square meter, what is the ideal radius and height of a cylindrical drum?

r = 2,85 feet

h = 8,56 feet

Step-by-step explanation:

Area of two circular caps (m ^(2)) : 2* pi* r^(2) (where  r is the radius of circular cap)     Area of wall (m ^(2))  is 2*pi*r*h (where h is the height of drum)

V= pi*r^(2)*h           so  h=V/pi*r^(2)   (1)

Cost in $:

Cost of wall 25 $/m^(2) plus 10% for manufacturing so:

C(w) = 27,5*2*pi*r*h $/square feet

Cost of (both top and bottom cap) = 2*40,97*pi* r^(2) so:

C(caps) = 257,29* r^(2) $/squaer feet

Total cost of cylnder = cost of wall + cost of caps

Total cost f cylinder: F(c)= 55*pi*r*h+257,29* r^(2)

F(c) = 172,7*r*h + 257,29*r^(2)         F(c)  is F(r)

Since from (1) we have h=218,276/pi*r^(2)  

F(r) = (172,7)*r*(69,514)/r^(2) +257,29*r^(2)           F(c) = 12005/r + 257,29*r^(2)

Taken the first dervative

F´(r)= -12005*(1)/r^(2) +2*257,29*r         F´(r) =  -12005*(1)/r^(2) +514,58*r

f F´(r) = 0          -12005*(1)/r^(2) + 514,58*r =0

-120005 + 514,58*^(3) = 0          514,58*^(3) = 12005      r =cubic root (23,32)

r = 2,85 ft

if we replace this value en F(r) we can see F(r) tend to infinite both when r tend to 0 and when r tends to infnite so there is a minimun for the functon

r = 2,85 ft    and   h = 8,56 ft

the expression (1.26)2t reveals the approximate rate of increases in the number of plants semiannually.

hope this : )

stops on an even number.