A rectangular piece of cardboard twice as long as it is wide is to be made into an open box by cutting 2-cm squares from each corner and bending up the sides. Express the volume V of the box as a function of the width w of the piece of cardboard.V(w) = 4w^2 - 24w + 32V(w) = 3w^2 + 12w + 23V(w) = 2w^2 - 24w + 20V(w) = w^2 - 50w + 11
Question
A rectangular piece of cardboard twice as long as it is wide is to be made into an open box by cutting 2-cm squares from each corner and bending up the sides. Express the volume V of the box as a function of the width w of the piece of cardboard.V(w) = 4w^2 - 24w + 32V(w) = 3w^2 + 12w + 23V(w) = 2w^2 - 24w + 20V(w) = w^2 - 50w + 11
Solution
The problem states that the length of the cardboard is twice its width. So, if we denote the width as w, then the length would be 2w.
When we cut a 2-cm square from each corner, we are reducing both the length and the width by 4 cm (2 cm from each side). So, the new dimensions of the cardboard when folded into a box would be:
- Length: 2w - 4 cm
- Width: w - 4 cm
- Height: 2 cm (the side of the square cut out)
The volume V of a rectangular box is given by the formula V = length * width * height. Substituting the new dimensions into this formula, we get:
V(w) = (2w - 4) * (w - 4) * 2
Expanding this out, we get:
V(w) = 4w^2 - 16w - 8w + 16
Simplifying, we get:
V(w) = 4w^2 - 24w + 16
So, the volume V of the box as a function of the width w of the piece of cardboard is V(w) = 4w^2 - 24w + 16. None of the options provided in the question match this result.
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