Posted by **diwas** on Tuesday, November 13, 2012 at 10:06pm.

a cabinet will inclose a rectangular loud speaker system and will have an internal volume of 2.4 cubic feet. The height will be 1.5 times the width. What are the dimesnions of the enclosure that can be constructed at a minimum cost?

- math -
**Reiny**, Tuesday, November 13, 2012 at 10:44pm
length --- x

width --- y

height ---- 1.5y

1.5xy^2 = 2.4

x = 2.4/(1.5y^2) = 8/(5y^2)

Assuming the cost is a function of the surface area (SA)

SA = 2xy + 2(1.5xy) + 2(1.5y^2)

= 2xy + 3xy + 3y^2

= 5y((8/(5y^2) + 3y^2

= 8/y + 3y^2

d(SA)/dy = -8/y^2 + 6y = 0 for a min of SA

8/y^2 = 6y

6y^3 = 8

y^3 = 8/6

y = 2/6^(1/3) = 1.1006.. = appr 1.1

1.5y = 1.65

x = 8/(5(1.1006..)^2) = 1.32

so

length = 1.32

width = 1.1

height = 1.65

check: V = 1.32x1.1x1.65 = appr 2.3958 , close to 2.4, not bad

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