Question 143854
Start by letting the width of the metal sheet be x inches, then the length is x+10 inches.
The volume of the box (832 cu.in.) will be calculated by the length (less 4 inches) times width (less 4 inches) times the height of the box (2 inches).
We can write the equation:
{{{V = (x-4)(x+10-4)(2)}}} or...
{{{(x-4)(x+6)(2) = 832}}} Divide both sides by 2 to simplify a bit.
{{{(x-4)(x+6) = 416}}} Perform the multiplication.
{{{x^2+2x-24 = 416}}} Subtract 416 from both sides.
{{{x^2+2x-440 = 0}}} Factor this quadratic equation.
{{{(x+22)(x-20) = 0}}} so that...
{{{x = -22}}} or {{{x = 20}}} Discard the negative solution as the length cannot be a negative quantity.
{{{x = 20}}}
The original dimensions of the metal sheet are:
Width = 20 inches and the length = 30 inches.
Check:
The volumes is:
(x-4)(x+6)(2)(2) = 832 Substitute x = 20.
{{{(20-4)(20+6)(2) = 832}}}
{{{(16)(26)(2) = 832}}}
{{{832 = 832}}} OK!