<PRE><B>Question 198229</B>
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Let &lt;b&gt;&lt;i&gt;w&lt;/i&gt;&lt;/b&gt; represent the width.  Then the length is &lt;b&gt;&lt;i&gt;w&lt;/i&gt; + 10&lt;/b&gt;.  Since both the length and the width are reduced by 5 &lt;i&gt;on each side&lt;/i&gt;, the width of the box becomes &lt;b&gt;&lt;i&gt;w&lt;/i&gt; - 10&lt;/b&gt; and the length of the box is &lt;b&gt;&lt;i&gt;w&lt;/i&gt; + 10 - 10&lt;/b&gt; or just &lt;b&gt;&lt;i&gt;w&lt;/i&gt;&lt;/b&gt;.  The height of the box, given the 5cm cutouts, must be 5cm.


The volume of the box, in cubic centimeters, is *[tex \LARGE 6L \times 1000cm^3/L = 6000cm^3].  And the volume of a rectangular solid is: *[tex \LARGE V = lwh], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5w(w-10) = 6000]


So, divide both sides by 5, distribute, put the equation into standard form, and solve the quadratic for *[tex \LARGE w] to obtain the original width.  Remember to exclude the negative root because you are looking for a positive measure.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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