document.write( "Question 621576: The length of a rectangular box is 1 inch more than twice the height of the box, and the width is 3 inches more than the height. If the volume of the box is 50 cubic inches, find the dimensions of the box. \n" ); document.write( "

Algebra.Com's Answer #391012 by ankor@dixie-net.com(22740) $\"\"$ $\"About$

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The length of a rectangular box is 1 inch more than twice the height of the box, and the width is 3 inches more than the height.
\n" ); document.write( " If the volume of the box is 50 cubic inches, find the dimensions
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\n" ); document.write( "Let x = the height of the box
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\n" ); document.write( "\"The length of a rectangular box is 1 inch more than twice the height\"
\n" ); document.write( "L = 2x+1
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\n" ); document.write( "\"width is 3 inches more than the height.\"
\n" ); document.write( "W = x+3
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\n" ); document.write( "L * W * H = 50
\n" ); document.write( "(2x+1)*(x+3)*x = 50
\n" ); document.write( "FOIL
\n" ); document.write( "(2x^2 + 6x + x + 3)*x = 50
\n" ); document.write( "x(2x^2 + 7x + 3) = 50
\n" ); document.write( "2x^3 + 7x^2 + 3x - 50 = 0
\n" ); document.write( "Plot this on your graphing calc,
\n" ); document.write( " $\"+graph%28+300%2C+200%2C+-5%2C+5%2C+-10%2C+10%2C+2x%5E3%2B7x%5E2%2B3x-50%29+\"$ \r
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\n" ); document.write( "\n" ); document.write( " see that the solution is x=2 is the height
\n" ); document.write( "then
\n" ); document.write( "2 cm = the height
\n" ); document.write( "2(2)+1 = 5 cm is the length
\n" ); document.write( "2 + 3 = 5 cm is the width\r
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