SOLUTION: An open box is formed by cutting a 5 inch square measured from each corner and folding up the sides. If the volume of the carton is then 40 in3, what was the length of a side of th

Algebra ->  Equations -> SOLUTION: An open box is formed by cutting a 5 inch square measured from each corner and folding up the sides. If the volume of the carton is then 40 in3, what was the length of a side of th      Log On


   

Question 1106241: An open box is formed by cutting a 5 inch square measured from each corner and folding up the sides. If the volume of the carton is then 40 in3, what was the length of a side of the original square of cardboard?
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
x, length of side of original square piece
5, edge length of each square piece removed, also same as box height
highlight_green%28%28x-2%2A5%29%28x-2%2A5%29%2A5=40%29
-
5%28x-10%29%28x-10%29=40
%28x-10%29%28x-10%29=8
x-10=0%2B-+2%2Asqrt%282%29
x=10%2B-+2%2Asqrt%282%29
highlight%28x=10%2B2%2Asqrt%282%29%29




MISREAD PROBLEM DESCRIPTION AND QUESTION***************************************************************
x, edge of each cut out square piece;
also the height of the open box
-
cross%28%285-2x%29%285-2x%29x=40%29
-
You might continue with the algebra (rational roots theorem,...); or you could look at some factorizations for 40 instead.
-
25x-20x%5E2%2B4x%5E3-40=0
4x%5E3-20x%5E2%2B25x-40=0
In case you prefer, you could try using a graphing tool.




-


Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
An open box is formed by cutting a 5 inch square measured from each corner and folding up the sides.
If the volume of the carton is then 40 in3, what was the length of a side of the original square of cardboard?
~~~~~~~~~~~~~~~~~~ 

Let x be "the length of a side of the original square of cardboard", which is under the question.


Then the dimensions of the bottom of the box are  x-2*5 = x - 10 inches.


Hence, the volume of the box is  5*(x-10)^2 cubic inches, and you have an equation

5*(x-10)^2 = 40.


It implies

(x-10)^2 = 40%2F5  ====>  (x-10)^2 = 8  ====>  x-10 = +/- sqrt%288%29 = +/- 2%2Asqrt%282%29  ====>  x%5B1%2C2%5D = 10 +- 2%2Asqrt%282%29.


But the value  (x-10) must be positive; it means that only   x= 10+%2B+2%2Asqrt%282%29  is the solution.


Answer.  The original cardboard  was the square of the side length  10+%2B+2%2Asqrt%282%29  inches.


-------------
The solution by @josgarithmetic is   W R O N G   and   I R R E L E V A N T.


Simply ignore it . . . for your safety . . .