We have 1000USD to buy the materials to build a box whose base length is seven times the base width and has no top. If the material for the sides costs 10USD/cm2 and the material of the bottom costs 15USD/cm2 determine the dimensions of the box that will maximise the enclosed volume.
Question from: http://tutorial.math.lamar.edu/ProblemsNS/CalcI/Optimization.aspx
I've tried solving it but each time I get a different answer and it doesn't make any sense. I got 164.65cm3 as an answer...
This is more a hint than a full solution! I can only help further if you can show what you are actually stuck on.
You can setup the problem as mathreadler suggested. Let $L$ be the length of the box, $H$ be the height of the box and $W$ the width of the box. Furthermore we define $c_b$ to be the cost of the bottom per unit area and $c_s$ the cost of the side per unit area.
Now we notice that $L = 7W$. We can now setup equations for the cost and the volume of the box.
The total volume, V, is given by: $$ V = L \times W \times H = 7W^2H $$
The total cost, C, is given by (where A denotes area): $$ C = A_{bottom}c_b + A_{side}c_s = 7W^2c_b + (W+7W+W+7W)Hc_s = 7W^2c_b + 16WHc_s$$
The total cost is constrained: $$ C = 7W^2c_b + 16WHc_s = 105W^2 + 160WH \leq 1000 $$
You now have to balance the width and hight of the box in such a way that the constraint is met, and the volume is as large as possible. You can do this using derivatives. Hopefully you can now do some sensible calculations