Optimization Problem

[yous13yous]

You are an engineer in charge of designing the dimensions of a box-like building. The base is rectangular in shape with width being twice as large as length. (Therefore so is the ceiling.) The volume is to be 36864000 m3. Local bylaws stipulate that the building must be no higher than 80 m. Suppose the walls cost twice as much per m2 as the ceiling, and suppose the floor (i.e.base) costs nothing. Find the dimensions of the building that would minimize the cost.

So far I have the following:
w=2l

V=36864000
l(2l)h=36864000
2l^2h=36864000
h=36864000/2l^2

x=cost of ceiling

so Total Cost=4x+x
TC=5x
and then I know I shoud find the derivative and then equate that to 0 but that gets me no where... :?

 

[Deleted member 4993]

You are an engineer in charge of designing the dimensions of a box-like building. The base is rectangular in shape with width being twice as large as length. (Therefore so is the ceiling.) The volume is to be 36864000 m3. Local bylaws stipulate that the building must be no higher than 80 m. Suppose the walls cost twice as much per m2 as the ceiling, and suppose the floor (i.e.base) costs nothing. Find the dimensions of the building that would minimize the cost.

So far I have the following:
w=2l

V=36864000
l(2l)h=36864000
2l^2h=36864000
h=36864000/2l^2

x=cost of ceiling

so Total Cost=4x+x
TC=5x
and then I know I shoud find the derivative and then equate that to 0 but that gets me no where... :?

Let

W = 2L

let cost/m[sup:15vtl0sw]2[/sup:15vtl0sw] = x for celing

Cost = x*[2{2(L+W)H} + L*W] = x*[12 * L *H + 2L[sup:15vtl0sw]2[/sup:15vtl0sw]] = 2*x*[6LH + L[sup:15vtl0sw]2[/sup:15vtl0sw]]

V(olume) = L*W*H

H = V/(2L[sup:15vtl0sw]2[/sup:15vtl0sw])

Then

Cost = 2*x*[6LH + L[sup:15vtl0sw]2[/sup:15vtl0sw]] = 2*x*[6L* V/(2L[sup:15vtl0sw]2[/sup:15vtl0sw]) + L[sup:15vtl0sw]2[/sup:15vtl0sw]] = 2*x*[ 3V/L + L[sup:15vtl0sw]2[/sup:15vtl0sw]] ]

Now differentiate and find minimum

Don't forget to check against maximum H allowable.

 

[yous13yous]

Let

W = 2L

let cost/m[sup:xuibq4wv]2[/sup:xuibq4wv] = x for celing

Cost = x*[2{2(L+W)H} + L*W] = x*[12 * L *H + 2L[sup:xuibq4wv]2[/sup:xuibq4wv]] = 2*x*[6LH + L[sup:xuibq4wv]2[/sup:xuibq4wv]]

V(olume) = L*W*H

H = V/(2L[sup:xuibq4wv]2[/sup:xuibq4wv])

Then

Cost = 2*x*[6LH + L[sup:xuibq4wv]2[/sup:xuibq4wv]] = 2*x*[6L* V/(2L[sup:xuibq4wv]2[/sup:xuibq4wv]) + L[sup:xuibq4wv]2[/sup:xuibq4wv]] = 2*x*[ 3V/L + L[sup:xuibq4wv]2[/sup:xuibq4wv]] ]

Now differentiate and find minimum

Don't forget to check against maximum H allowable.[/quote]

But there are 2 variables: x and L

 

[Deleted member 4993]

For this problem 'x' is not a variable - only 'L' is variable.

 

[yous13yous]

For this problem 'x' is not a variable - only 'L' is variable.

Lets just say my derivative is
C'=2(-3VL^-2 + 2L)
C'=-6VL^-2 + 4L
0=-211184000L^-2 + 4L
211184000L^-2=4L
55296000L^-3=1
L^3=55296000
L=380.98 m

and so
W=2(380.98) = 761.95 m

and then V=LWH
so H=V/(LH)
H=36864000/(380.98*761.95) = 126.99 m

Which is totally contradictory because 0<=H<=80
so I dont know where I went wrong

 

[Deleted member 4993]

You did not go wrong anywhere.

So now you know the H = 80 m

Now calculate L & W knowing V and constraint that W = 2L

 

[yous13yous]

You did not go wrong anywhere.

So now you know the H = 80 m

Now calculate L & W knowing V and constraint that W = 2L

Ok so since
V=HLW
36864000=80(2L^2)
36864000=160L^2
L^2=230400
L=480

and since W=2L=2(480)=960

Therefore the 3 dimensions are
Length: 480 m
Width: 960 m
Height: 80 m

THANK YOU SOOOO MUCH!!!