SOLUTION: 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 yo

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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.
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
The cost equation is
C=20D%2B30W%2B50H
You also know that
V=D%2AW%2AH=600
You can replace one of the variables in the cost equation.
D=600%2F%28WH%29
Then the cost equation becomes,
C=20%28600%2F%28WH%29%29%2B30W%2B50H
C=%2812000%29%2F%28WH%29%2B30W%2B50H
C=12000W%5E%28-1%29H%5E%28-1%29%2B30W%2B50H
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%2FdW=12000H%5E%28-1%29%28-W%5E%28-2%29%29%2B30=0
2.dC%2FdH=12000W%5E%28-1%29%28-H%5E%28-2%29%29%2B50=0
The understanding is that dC%2FdW and dC%2FdH are partial derivatives.
From eq. 1,
12000H%5E%28-1%29%28-W%5E%28-2%29%29%2B30=0
12000H%5E%28-1%29%28W%5E%28-2%29%29=30
3.HW%5E2=400
From eq. 2,
12000W%5E%28-1%29%28-H%5E%28-2%29%29%2B50=0
12000W%5E%28-1%29%28H%5E%28-2%29%29=50
4.H%5E2W=240
From eq. 4,
4.H%5E2W=240
W=240%2FH%5E2
W%5E2=57600%2FH%5E4
Substitute this expression in eq. 3 and solve for H,
3.HW%5E2=400
H%2857600%2FH%5E4%29=400
57600%2FH%5E3=400
H%5E3=57600%2F400=144
H=5.24
From eq. 4,
W=240%2FH%5E2
W=240%2F%285.24%29%5E2
W=240%2F%2827.4576%29
W=8.74
Finally,
D=600%2F%28WH%29
D=600%2F%285.24%2A8.74%29
D=600%2F45.7976
D=13.10
The total cost would then be,
C=20%2813.10%29%2B30%2A%288.74%29%2B50%2A%285.24%29
C=262%2B262.2%2B262.... An interesting finding, minimizing cost means each portion makes up exactly 1/3 of the cost.
C=786.2