SOLUTION: A rectangular piece of sheet metal has a length that is 2 times its width. In each corner a square with sides of 2.4 inches is cut out. The outer strips are then bent up to fr

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: A rectangular piece of sheet metal has a length that is 2 times its width. In each corner a square with sides of 2.4 inches is cut out. The outer strips are then bent up to fr      Log On


   

Question 958124:
A rectangular piece of sheet metal has a length that is 2 times its width. In each corner a square with sides of 2.4 inches is cut out. The outer strips are then bent up to from an open box with a volume of 155
in3
. Find the length (in inches) of the original sheet of metal.
Express your answer to three significant digits

Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Best to do this completely in variables and substitute the given values last.

u, the side length of the cut out squares;
w, original width
L, original length
v, volume of the open box formed
-
L=2w
u=2.4
v=155
-
Goal is to solve for L and w.

Folding the flaps makes a base area of the box, %28w-2u%29%28L-2u%29=%28w-2u%29%282w-2u%29.

The volume will be the base area multiplied by u.
%28w-2u%29%282w-2u%29u=v
This is an equation with only one unknown, being w. The other variables are given.

Simplify that equation.
%282w%5E2-4uw-2uw%2B4u%5E2%29u-v=0
2uw%5E2-4u%5E2w-2u%5E2w%2B4u%5E3-v=0
highlight_green%282uw%5E2-5u%5E2w%2B4u%5E3-v=0%29

Instead of continuing this completely in symbols, substituting the given values NOW might make the rest of the work, either factoring if possible, or general solution for a quadratic equation, easier to do. Your choice. You take the rest of this. Solve for w, and use the result to evaluate L.