Arectangular piece of metal is 10 in. longer than it is wide. squares with sides 2 in. long are cut from the four corners

3 answers
Question:

Arectangular piece of metal is 10 in. longer than it is wide. squares with sides 2 in. long are cut from the four corners and the flaps are folded upward to form an open box. if the volume of the box is 832 in. cubed, what were the original dimensions of the piece of metal

Answers

Volume of the box = length  * width * height

length of original peice = x + 10 , width = x

length of box  (x + 10) - 4  = x + 6
width of box = x - 4
height of box =  2

so we have 

2 * (x -4) * ( x + 6) = 832

2x^2 + 4x - 48 = 832

2x^2 + 4x - 880 = 0
x^2  + 2x - 440 = 0

(x + 22)(x - 20) = 0

x = 20

so width of original piece of metal = 20 ins , length = 30 inches.

The dimensions of the piece of metal can be represented by x and x+10.

 

Now, 2 in squares are being cut out of each corner.

 

So the new dimensions are x-4 (2in from each side) and x+10-4 or x+6.

 

When you fold it up, the height becomes 2 and the base has dimensions x-4 and x+6. Now plug this into the volume formula.

 

V=l*w*h

 

1302 = (x+6)(x-4)(2)

651   = x2+2x-24

0      = x2 + 2x-675

0      = (x+27)(x-25)

x=-27 reject since lengths cannot be negative or x=25.

 

So your original dimension for the piece of metal are 25 by 35.  

 

Hope this helped.

Step-by-step explanation:

The dimensions of the piece of metal can be represented by x and x+10.

Now, 2 in squares are being cut out of each corner.

So the new dimensions are x-4 (2in from each side) and x+10-4 or x+6.

When you fold it up, the height becomes 2 and the base has dimensions x-4 and x+6. Now plug this into the volume formula.

V=l*w*h

1302 = (x+6)(x-4)(2)

651   = x2+2x-24

0      = x2 + 2x-675

0      = (x+27)(x-25)

x=-27 reject since lengths cannot be negative or x=25.

So your original dimension for the piece of metal are 25 by 35.

Step-by-step explanation:

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