1. ## Minimization, finding dimensions

The cabinet that will enclose the Acrosonic model D loudspeaker system will be rectangular and will have an internal volume of 1.8 ft3. For aesthetic reasons, it has been decided that the height of the cabinet is to be 1.4 times its width. If the top, bottom, and sides of the cabinet are constructed of veneer costing 35¢ per square foot and the front (ignore the cutouts in the baffle) and rear are constructed of particle board costing 19¢ per square foot, what are the dimensions of the enclosure that can be constructed at a minimum cost?

This is a minimization problem. Is this a minimization of area or volume? What I have so far.

V=LWH. 1.8 = x^2(1.4x) my constraint? I’m assuming the bottom is a square so LW is both x and H is 1.4x.
Cost = 2bottoms * .35 + 2side *.35 + 2side *.19 = .35(2x^2) + .35((2(x*1.4x)) +.19((2(x*1.4x))

This problem is a little different than my previous minimization maximization problems. I can’t seem to combine them to find a derivative. Please help thank you!

2. Hi. I did this quick so check me ok:

$\displaystyle v=Lwh=1.8$

$\displaystyle C_t=V+P$

$\displaystyle V=\frac{35}{100}(t+b+2s)$

$\displaystyle P=\frac{19}{100}(f+r)$

Then I get:

$\displaystyle C_t=\left\{\frac{35}{100}(Lw+Lw+2hw)+\frac{19}{100 }(2hL)\right\}$

but $\displaystyle h=1.4 w$ and $\displaystyle L=\frac{1.8}{1.4 w^2}$

when I put all that into the cost function as a function of w I get:

$\displaystyle C_t=\left\{\frac{35}{100}(2\frac{1.8}{1.4 w})+2(1.4 w^2)+\frac{19}{100}(2(\frac{1.8}{w}))\right\}$

Next thing I would do is outright plot it and see if it has a minimum. If the plot doesn't, then I made a mistake and no need to go further. I'll leave that for you to check. If it does, then find the extrema to find the minimum cost.

3. Thank you Shawsend, I will re-try the problem with your help.