document.write( "Question 1026674: Pls, help me with this question
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document.write( "A farmer has 120m of fencing with which to enclose a rectangular sheep-pen, using a wall for one side. Find the maximum area that he can enclose \n" );
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Algebra.Com's Answer #641962 by mathmate(429)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "Question: \n" ); document.write( "Pls, help me with this question \n" ); document.write( "A farmer has 120m of fencing with which to enclose a rectangular sheep-pen, using a wall for one side. Find the maximum area that he can enclose \n" ); document.write( " \n" ); document.write( "Solution: \n" ); document.write( "This is a calculus question, but there does not seem to be a category for it. \n" ); document.write( "The perimeter of the pen is 120=w+w+(120-2w), where w is the width and 120-w is the length opposite the wall. \n" ); document.write( "The corresponding area is A(w)=w(120-2w)=120w-2w^2. \n" ); document.write( "By completing squares, \n" ); document.write( "A(w)=-2(w-30)^2+1800 \n" ); document.write( "We can see that when the first term vanishes, i.e. w=30, A(w) will be at its maximum. \n" ); document.write( "So the maximum area is \n" ); document.write( "A(30)=120*30-2*30^2=900, or w=30, L=120-2w=60. \n" ); document.write( "Erratum: \n" ); document.write( "The last line should read: \n" ); document.write( "A(30)=120*30-2*30^2=3600-1800=1800, or w=30, L=120-2w=60, and area=1800 |