An open box contains 80cm^3 and is made from a square piece of tinplate with 3cm squares cut from each of its four corners. Find the dimensions of the original piece of tinplate.
A=80
A=(length)(width)(height)
since length and width are the same, and when the box is folded the height will be 3cm,
80=(w^2)(3)
Since the width has 2 3cm sections cut out of it, I said:
w=x+6
To find x, I tried subbing w=x+6 into my previous equation:
80=((x+6)^2)*3
Which expands as:
(3x^2)+(36x)+28=0
What do I do next? I tried the quadratic formula to get x, but ended up with two negative numbers, which is not a valid solution. Please help me... Was my process wrong?
I posted this a few hours ago but I wasn't answered.
3 answers