MATH JOURNAL 19.2.2007
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19 Feb 2007, 10:55
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Journals
Minimizing the Cost of a Container
A trash company is designing an open-top, rectangular container that will have a volume of 320 m^3. The cost of making the bottom of the container is $5 per square m, and the cost of the sides is $4 per square m. Find the dimensions of the container that will minimize the total cost.
discuss...
and answer ;*
A trash company is designing an open-top, rectangular container that will have a volume of 320 m^3. The cost of making the bottom of the container is $5 per square m, and the cost of the sides is $4 per square m. Find the dimensions of the container that will minimize the total cost.
discuss...
and answer ;*
make a code:
"min 5x+4y
subject to
x*y=320
end"
V=xyz=V0 => xyz-V0 = 0
S=xy+2xz+2yz=f(x,y,z)
f must be minimum. This is a classsical equation with "connections". Use the Lagrange method to solve it. IF u dont know how ill try to solve it for u.
PS: Or u can take z from the first equation and put in the second one and use the simple derivation way.
equotion was 5xy+8xz+8yz
x=8 y=8 z=5 ;s
f(x,y,z) = 5xy+8xz+8yz
f(x, y) = 5xy + 8V0(1/x+1/y) (i took z from the first and put it in f).
df/dx = 5y-8V0(1/x^2) = 0
df/dy = 5x-8V0(1/y^2) = 0
so..
5xy^2 = 5yx^2 = 8V0 => x = y = (V0*8/5)^1/3 ( i think).
z = V0/xy.
Problem solved i think.
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