document.write( "Question 68092This question is from textbook Algebra and Trigonometry
\n" ); document.write( ": A rancher plans to use 160 yards of fencing to enclose a rectangular corral and to divide it into two parts by a fence parallel to the shorter sides of the corral. Find the dimensions of the corral if its are is 100yd2. \n" ); document.write( "
Algebra.Com's Answer #48551 by ankor@dixie-net.com(22740)\"\" \"About 
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A rancher plans to use 160 yards of fencing to enclose a rectangular corral and to divide it into two parts by a fence parallel to the shorter sides of the corral. Find the dimensions of the corral if its area is 100 yd^2.
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\n" ); document.write( "As written this problem does not make sense, that much fence would have a
\n" ); document.write( "much larger area. I'm going to assume you meant 1000 sq yds
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\n" ); document.write( "Perimeter in this case would be
\n" ); document.write( "2L + 3W = 160
\n" ); document.write( "2L = 160 - 3W
\n" ); document.write( "L = 160/2 - (3/2)W
\n" ); document.write( "L = 80 - 1.5W
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\n" ); document.write( "The area:
\n" ); document.write( "L*W = 1000
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\n" ); document.write( "Substitute (80-1.5W) for L in the area equation:
\n" ); document.write( "(80-1.5W) * W = 1000
\n" ); document.write( ":
\n" ); document.write( "80W - 1.5W^2 = 1000
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\n" ); document.write( "Arrange as a quadratic equation:
\n" ); document.write( "-1.5W^2 + 80W - 1000 = 0
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\n" ); document.write( "Using the quadratic formula, solutions are W=20 and W=35 which make sense:
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\n" ); document.write( "Using a width of 20 yds would make the length: 80 - 1.5(20) = 50 yds
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\n" ); document.write( "This checks out: 2(50) + 3(20) = 160
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