document.write( "Question 304826: A rectangular field is to enclosed with 600 m of fencing. What dimensions will produce a maximum area? \n" ); document.write( "
Algebra.Com's Answer #218296 by Edwin McCravy(20054)\"\" \"About 
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A rectangular field is to enclosed with 600 m of fencing. What dimensions will produce a maximum area?
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document.write( "Area = length * width\r\n" );
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document.write( "Let y = the area\r\n" );
document.write( "Let x = the width\r\n" );
document.write( "Let L = the length\r\n" );
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document.write( "      y = x * L\r\n" );
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document.write( "Now since the perimeter is 600\r\n" );
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document.write( "      P = 2*length + 2*width\r\n" );
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document.write( "    600 = 2L + 2x\r\n" );
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document.write( "Divide through by 2\r\n" );
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document.write( "    300 = L + x\r\n" );
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document.write( "Solve for L by subtracting x from both sides:\r\n" );
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document.write( "300 - x = L\r\n" );
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document.write( "Substituting 300-x for L in\r\n" );
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document.write( "      y = x * L\r\n" );
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document.write( "      y = x * (300 - x)\r\n" );
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document.write( "      y = 300x - x2\r\n" );
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document.write( "Write in descending order of exponents\r\n" );
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document.write( "      y = -x2 + 300x \r\n" );
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document.write( "Since the coefficient of \"x%5E2\" is negative,\r\n" );
document.write( "the parabola opens downward and has a maximum\r\n" );
document.write( "value at the vertex.  So we use the vertex\r\n" );
document.write( "formula:\r\n" );
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document.write( "The vertex of the parabola whose equation is \r\n" );
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document.write( "      y = ax2 + bx + c\r\n" );
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document.write( "has for its x coordinate \"-b%2F%282a%29\"\r\n" );
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document.write( "and then its y-coordinate is found by substituting\r\n" );
document.write( "that value for x in the equation and solving for\r\n" );
document.write( "y.\r\n" );
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document.write( "Write your equation as\r\n" );
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document.write( "      y = -1x2 + 300x + 0\r\n" );
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document.write( "And a=-1, b=300 and c=0\r\n" );
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document.write( "So the vertex of the parabola whose equation is \r\n" );
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document.write( "      y = -1x2 + 300x + 0\r\n" );
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document.write( "has for its x coordinate \"-b%2F%282a%29=-%28300%29%2F%282%2A%28-1%29%29=+150\"\r\n" );
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document.write( "That's the width of the rectangle of maximum area.  We need to\r\n" );
document.write( "find the length L from the equation above\r\n" );
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document.write( "300 - x = L\r\n" );
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document.write( "300 - 150 = L\r\n" );
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document.write( "150 = L\r\n" );
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document.write( "So we see that the largest area possible is when the rectangle\r\n" );
document.write( "is chosen to be the square 150 m by 150 m\r\n" );
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document.write( "You weren't asked for the value of that maximum area. But if\r\n" );
document.write( "you had been, then the y-coordinate of the vertex would have\r\n" );
document.write( "given us that by substituting 150 for x in the equation and \r\n" );
document.write( "solving for y.\r\n" );
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document.write( "      y = -1x2 + 300x + 0\r\n" );
document.write( "      y = -1(150)2 + 300(150) + 0\r\n" );
document.write( "      y = -1(22500) + 45000 + 0 \r\n" );
document.write( "      y = -22500 + 45000\r\n" );
document.write( "      y = 22500\r\n" );
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document.write( "So that maximum area would be 22500 square meters.\r\n" );
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document.write( "The graph is:\r\n" );
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document.write( "Edwin
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