Question
A 12cm by 8cm rectangular piece of metal is to be made into an open-top box by cutting a sqaure from corner and folding up the resulting flaps (sides). If the volume of the lidless box is 36 cm what are the integer dimensions of the box?
Answers
Reiny
Square to be cut out : x cm by x cm
then the height of the box is x, and the base is
12-2x by 8-2x, where 0 < x < 4
volume = x(12-2x)(8-2x)
= 36
x(6-x)(4-x) = 9
x(24 - 10x + x^2) = 9
x^3 - 10x^2 + 24x - 9 = 0
I tried x = ±1, ±3 and x = 3 was a solution.
by synthetic division, I got
(x-3)(x^2 - 7x + 3) = 0
the quadratic has no real solutions, so
x = 3 and the box will be
6 by 2 by 3
then the height of the box is x, and the base is
12-2x by 8-2x, where 0 < x < 4
volume = x(12-2x)(8-2x)
= 36
x(6-x)(4-x) = 9
x(24 - 10x + x^2) = 9
x^3 - 10x^2 + 24x - 9 = 0
I tried x = ±1, ±3 and x = 3 was a solution.
by synthetic division, I got
(x-3)(x^2 - 7x + 3) = 0
the quadratic has no real solutions, so
x = 3 and the box will be
6 by 2 by 3
aman
hi,
how did you go from 36=x(12-2x)(8-2x) to 9=x(6-x)(4-x)?
how did you go from 36=x(12-2x)(8-2x) to 9=x(6-x)(4-x)?