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garik1379 [7]
3 years ago
11

A standard oil drum is a cylinder with a volume of approximately 218.276 liters (or 0.218276 cubic meters). The cost of producin

g 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?
Mathematics
1 answer:
Arte-miy333 [17]3 years ago
5 0

Answer:

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

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Rewrite as perfect squares

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