1. ## Optimization problem.

The length of a cedar chest is twice its width. The cost/dm² of the lid is 4 times the cost/dm² of the rest of the cedar chest. If the volume of the cedar chest is 1440 dm³, find the dimensions so that the cost is a minimum.

Unfortunately, I don't know where to start (didn't see any questions like this at all in my examples or the questions I've solved).

2. Start by finding a function to represent the cost of making the chest...

3. Originally Posted by Prove It
Start by finding a function to represent the cost of making the chest...
How would I do that? That seems to be my only problem at the moment. Once I can do that, I can pretty much do the rest (differentiate, find where the derivative is 0, then plug back the minimum into the original equations to find the other unknowns). I know that LWH = 1440³ dm, and that L = 2W. Plugging this back into the equation I get 2W²H = 1440 and then I rearrange for $H = \frac{720}{W^{2}}$ but that's about as far as I can get.

4. It should be pretty clear that the cost is the same as the surface area of an open box, plus 4 times that cost for the lid...

5. Originally Posted by Prove It
It should be pretty clear that the cost is the same as the surface area of an open box, plus 4 times that cost for the lid...
So, $C = 2LW+2LH+2WH+4LW$?

6. No, first evaluate the surface area for the OPEN box (i.e. without the lid...).

Then the lid will be 4 times all of that...

7. Originally Posted by Prove It
No, first evaluate the surface area for the OPEN box (i.e. without the lid...).

Then the lid will be 4 times all of that...
$C = 4(2LH + 2WH + 2LW)$?

Thanks for all the help, by the way.

8. No.

Bottom: $\displaystyle LW$

Sides: $\displaystyle 2LH + 2WH$

4 times that: $\displaystyle 4(LW + 2LH + 2WH)$.

So total cost: $\displaystyle LW + 2LH + 2WH + 4(LW + 2LH + 2WH)$.

Now simplify and substitute your other formulas to make a function just of $\displaystyle L$ or $\displaystyle W$.