SOLUTION: I NEED HELP - OH PLEASE!!!I need to determine the outside dimensions of a large container. some of the info, which I am not sure is needed...they will need 5.15 cubic feet of stee

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: I NEED HELP - OH PLEASE!!!I need to determine the outside dimensions of a large container. some of the info, which I am not sure is needed...they will need 5.15 cubic feet of stee      Log On


   

Question 95160: I NEED HELP - OH PLEASE!!!I need to determine the outside dimensions of a large container. some of the info, which I am not sure is needed...they will need 5.15 cubic feet of steel, which will make the tank 1/2 inch thick...the width is 1 foot less than the length, the height is 9 feet more than the length. Thanks for the help!
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Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
5.15 ft³ at 1/2" thick is 5.15/(1/24) or 123.6 ft² of steel ... this is the approximate surface area

let x=length, so x-1=width and x+9=height ... SA=2((x)(x-1)+(x)(x+9)+(x-1)(x+9))

SA=2((x^2-x)+(x^2+9x)+(x^2+8x-9))=2(3x^2+16x-9) ... 123.6=6x^2+32x-18 ... 0=6x^2+32x-141.6

quadratic formula gives x=%28-32+%2B-+sqrt%2832%5E2-4%286%29%28-141.6%29%29%29%2F%282%286%29%29 ... x=2.875 and x=-8.208 ... negative value not realistic

length=2.875, width=1.875, height=11.875

NOTE: 123.6 ft² is the amount of plate used ... the actual SA will depend on how the corners are joined