## anonymous 5 years ago A box with a rectangular base is to be constructed of material costing $2/inch^2 for the sides and the bottom, and$3/inch^2 for the top. If the volum is 1215 inches^3 and the length is twice the width, which dimensions minimize cost. I understand that I have "l=2w" and "l*w*h=1215" so "3w*h=1215" but I don't know how to eliminate h or w.

1. anonymous

You have a box with the following dimensions: length = 2w width = w height = h The cost of the material will depend upon how much of each material we use. According to your specifications, we have for the surface area of the bottom and sides,$A_{bottom,sides}=2w^2+2(2wh)+2(wh)=2w^2+6wh$and$A_{t{op}}=2w^2$The total cost will then be$c=2.A_{bottom,sides}+3.A_{t{op}}$That is,$c=12wh+10w^2$Now, you had a problem with finding the height. You're given another condition in your problem statement, that of the volume being specified. We can find the height in terms of w from the formula for the volume$V=2w^2h \rightarrow h=\frac{V}{2w^2}$Substituting this in for h in the cost function, you have$c=\frac{6V}{w}+10w^2$From here it's just standard - find the derivative of c with respect to w, set it to zero, solve for w and substitute that back into c(w) to find the minimal cost. You should (technically speaking) find the second derivative of c(w) also, to show that it is positive, implying the solution for w will determine a c(w) that is minimal. The w you should end up with (assuming my scratching's right) is$w=\left( {\frac{3V}{10}} \right)^{\frac{1}{3}}\approx 7.14inches$

2. anonymous

Your dimensions and cost can now be found.

3. anonymous

Wow! Lifesaver! I'm your biggest fan, thank you very much.

4. anonymous

Cheers :) Just check everything, though. I was half asleep when I did it.

5. anonymous

You came out with the correct answer and all the math seems to fit. Again thanks!