SOLUTION: An open metal chemical tank is to be made from a rectangular piece of stainless steel that measures 10 feet by 8 feet, by cutting out squares of the same size from each corner and

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Question 130510This question is from textbook College Algebra with Trigonometry
: An open metal chemical tank is to be made from a rectangular piece of stainless steel that measures 10 feet by 8 feet, by cutting out squares of the same size from each corner and bending up the sides. If the volume of the tank is to be 48 cubic feet, how large a square should be cut from each corner?
The problem asks to find all rational solutions exactly, and find irrational solutions to one decimal place.
I am truly confused from this problem! I have done others like it, but my teacher usually gets it started for us. Any help whatsoever is appreciated very very much!!
This question is from textbook College Algebra with Trigonometry

Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
the side of the square will be the depth of the tank
__ the length and width of the tank will be the the length, 10ft, and width, 8ft, of the sheet
__ each dimension reduced by twice the side of the square - a square is cut from each end

the volume is length times width times depth __ let x="side of square"
__ so volume=(10-2x)(8-2x)(x) __ 48=80x-36x^2+4x^3

4x^3-36x^2+80x-48=0 __ dividing by 4 __ x^3-9x^2+20x-12=0

possible values for x are the factors of -12 __ 1, -1, 2, -2, 3, -3, 4, -4, 6, -6, 12, -12
__ 1 satisfies the equation, so x-1 can be factored out

factoring __ (x-1)(x^2-8x+12)=0 __ factoring again __ (x-1)(x-2)(x-6)=0

1, 2, and 6 all satisfy the equation
__ but in real life, you can't cut a 6ft square from all 4 corners of an 8ft by 10ft sheet