Asked by Azka
From each corner of a square piece of cardboard, a square with sides of 3cm is removed. The edges are then up to form an open box. If the box is to hold 243cm^3, what are the dimensions of the original piece of cardboard?
Answers
Answered by
Reiny
original piece ---- x cm by x cm
base is (x-6) by (x-6)
x(x-6)^2 = 243
x^3 - 12x^2 + 36x - 243 = 0
I don't know what method you have to solve a cubic. Since this one does not factor it becomes complicated. Newton's Method is one algorithm.
I sent it through Wolfram and got 10.754
http://www.wolframalpha.com/input/?i=x%5E3+-+12x%5E2+%2B+36x+-+243+%3D+0
base is (x-6) by (x-6)
x(x-6)^2 = 243
x^3 - 12x^2 + 36x - 243 = 0
I don't know what method you have to solve a cubic. Since this one does not factor it becomes complicated. Newton's Method is one algorithm.
I sent it through Wolfram and got 10.754
http://www.wolframalpha.com/input/?i=x%5E3+-+12x%5E2+%2B+36x+-+243+%3D+0
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