Problem involving polynomial equations

[Byrgg]

Identical squares are cut out from each corner of a rectangular sheet of tin 8cm x 6cm. The sides are bent upward to form an open box. If the volume of the box is 16cm^3, what is the length of each side of the squares cut out from the original sheet?

I came up with the following equations to start, I don't know if they're right though:

xyz = 16cm^3
2x + y = 8
2x + z = 6

I'll try to make a diagram and put it up later, if anyone thinks it would help.

 

[dmoravec]

The equations look good. If you can eliminate either the y or the z from one of the second two equations you can solve it easily.

 

[Architeuthis Dux]

Your equations are a good start, but they are not quite accurate. The first equation should be (x-2y)(y-2x)z = 16cm^3, since the volume of the box is equal to the product of its dimensions. The second and third equations should be (x-2y) = 8 and (y-2x) = 6, since the sides of the box are formed by bending the sides of the original sheet.

To solve this problem, we can use the substitution method. From the second equation, we can solve for y in terms of x, giving us y = (8-2x)/2 = 4-x. We can then substitute this value of y into the first equation to get (x-2(4-x))(4-x-2x)z = 16cm^3. Simplifying this, we get (x-8)(-3x+4)z = 16cm^3. We can also substitute the value of y into the third equation to get (4-x-2x) = 6, which simplifies to -3x = 2, or x = -2/3.

Substituting this value of x into our first equation, we get (-2/3-8)(-3(-2/3)+4)z = 16cm^3. Simplifying this, we get (8/3)(10/3)z = 16cm^3, or z = 9/5. Now we can substitute these values of x and z into our second equation to solve for y. We get (x-2y) = 8, or (-2/3-2y) = 8, or y = -17/6.

Therefore, the length of each side of the squares cut out from the original sheet is -2/3 cm. We can check this solution by plugging these values into our original equations and seeing if they satisfy the conditions. We can also graph the system of equations and see where they intersect, which should give us the same solution.