Questions on Algebra: Linear Algebra (NOT Linear Equations) answered by real tutors!

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Question 164817: I have to find the least expensive cost for a steel frame building which measures 600 cu. meters. The prices of horizontal beams are 20d and 30w, vertical beams are 50h. Thank you.: I have to find the least expensive cost for a steel frame building which measures 600 cu. meters. The prices of horizontal beams are 20d and 30w, vertical beams are 50h. Thank you.
Answer by Fombitz(1799) About Me  (Show Source):
You can put this solution on YOUR website!
The cost equation is
C=20D+30W+50H
You also know that
V=D*W*H=600
You can replace one of the variables in the cost equation.
D=600/(WH)
Then the cost equation becomes,
C=20(600/(WH))+30W+50H
C=(12000)/(WH)+30W+50H
C=12000W^(-1)H^(-1)+30W+50H
To minimize the cost function we have to find the partial derivative of C with respect to W and H and set those equal to zero.
1.dC/dW=12000H^(-1)(-W^(-2))+30=0
2.dC/dH=12000W^(-1)(-H^(-2))+50=0
The understanding is that dC/dW and dC/dH are partial derivatives.
From eq. 1,
12000H^(-1)(-W^(-2))+30=0
12000H^(-1)(W^(-2))=30
3.HW^2=400
From eq. 2,
12000W^(-1)(-H^(-2))+50=0
12000W^(-1)(H^(-2))=50
4.H^2W=240
From eq. 4,
4.H^2W=240
W=240/H^2
W^2=57600/H^4
Substitute this expression in eq. 3 and solve for H,
3.HW^2=400
H(57600/H^4)=400
57600/H^3=400
H^3=57600/400=144
H=5.24
From eq. 4,
W=240/H^2
W=240/(5.24)^2
W=240/(27.4576)
W=8.74
Finally,
D=600/(WH)
D=600/(5.24*8.74)
D=600/45.7976
D=13.10
The total cost would then be,
C=20(13.10)+30*(8.74)+50*(5.24)
C=262+262.2+262.... An interesting finding, minimizing cost means each portion makes up exactly 1/3 of the cost.
C=786.2