A rectangular sheet of cardboard has a width of `h` centimetres. Its length is twice its width.
Squares of side length `x` centimetres, where `x > 0` are cut from each of the corners, as shown in the diagram below.
The sides of this sheet of cardboard are then folded up to make a rectangular box with an open top, as shown in the diagram below.
Assume that the thickness of the cardboard is negligible and that `V_text{box} > 0`.
A box is to be made from a sheet of cardboard with `h` = 25 cm.
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- Show that the volume, `V_text{box} > 0`, in cubic centimetres, is given by `V_text{box} (x) = 2x (25 - 2x)(25 - x)`. (1 mark)
- State the domain of `V_text{box}`. (1 mark)
- Find the derivative of `V_text{box}` with respect to `x`. (1 mark)
- Calculate the maximum possible volume of the box and for which value of `x` this occurs. (3 marks)
- Waste minimisation is a goal when making cardboard boxes.
- Percentage wasted is based on the area of the sheet of cardboard that is cut out before the box is made.
- Find the percentage of the sheet of cardboard that is wasted when `x = 5`. (2 marks)
Now consider a box made from a rectangular sheet of cardboard where `h>0` and the box's length is still twice its width.
- i. Let `V_text{box}` be the function that gives the volume of the box.
- State the domain `V_text{box}` in terms of `h`. (1 mark)
- ii. Find the maximum volume for any such rectangular box, `V_text{box}`, in terms of `h`. (3 marks)
- Now consider making a box from a square sheet of cardboard with side lengths of `h` centimetres.
- Show that the maximum volume of the box occurs when `x = h/6`. (2 marks)