Question 1134803
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Let the dimensions of the original piece of metal be x and x+30.<br>
When squares of side length 6 are cut from each corner to form the open box, the dimensions of the base are x-12 and (x+30)-2 = x+18; the height of the box is 6.<br>
We are told that the volume of the box (length times width times height) is 816:<br>
{{{6(x-12)(x+18) = 816}}}
{{{(x-12)(x+18) = 136}}}
{{{x^2+6x-216 = 136}}}
{{{x^2+6x-352 = 0}}}
{{{(x+22)(x-16) = 0}}}
{{{x = -22}}} or {{{x = 16}}}<br>
Reject the negative answer; it makes no sense in the problem.<br>
ANSWER: The dimensions of the original piece of metal were x=16 and x+30 = 46.