the cabinet that will enclose the Acrosonic model D loudspeaker system will be r

Question

the cabinet that will enclose the Acrosonic model D loudspeaker system will be rectangular and will have an internal volume of 2.4 ft^3. For aesthetic reasons, the design team has decided that the height of the cabinet is to be 1.5 times its width. if the top, bottom, and sides of the cabinet are constructed of veneer costing 40 cents a square foot and the fount (ignore the cutouts in the baffle) and rear are constructed of particle board costing 20 cents a square foot, what are the dimensions of the enclosure that can be constructed at the minimum cost?

Explanation / Answer

V = 2.4
h = 1.5w

V = L * w * h
2.4 = L * w * (1.5w)
L = 2.4 / (1.5w^2)

A = 2Lw + 2Lh + 2wh

Uh... Question... which side is the front/back? >.< Lh or wh?

*assuming wh is the front/back*

C = 40(2Lw + 2Lh) + 20(2wh)
C = 80(Lw + Lh) + 40(wh)
C = 80( 2.4 / (1.5w^2) * w + 2.4 / (1.5w^2) * 1.5w) + 40(w * 1.5w)
C = 80( 2.4 / (1.5w) + 2.4 / w) + 40(1.5w^2)
C = 192( 1/(1.5w) + 1/w) + 60w^2
C = 128/w + 192/w + 60w^2
C = 60w^2 + 320/w
C' = 120w - 320/w^2
320/w^2 = 120w
320 = 120w^3
w^3 = 8/3
w = 1.387 ft.
h = 2.080 ft.
L = 0.832 ft.

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