document.write( "Question 55951: Hello, I would like to know if I did this problem correctly. Can you help me, please?
\n" ); document.write( "Amanda has 400 feet of lumber to frame a rectangular patio(the perimeter of a rectangle is 2 times length plus 2 times width). What should the dimensions of the patio be, and show how the maximum area of the patio is calculated from the algebraic equation. Use the vertex form to find the maximum area.
\n" ); document.write( "2L = 400 - 2W
\n" ); document.write( "A=LW
\n" ); document.write( "=(400-2W)W
\n" ); document.write( "=4==W-2W^\r
\n" ); document.write( "\n" ); document.write( "A= 2W^ - 400W
\n" ); document.write( "W = 400/2(-2) = 200 FT
\n" ); document.write( "L = 400 - 2W - 400 2(200)
\n" ); document.write( "=400\r
\n" ); document.write( "\n" ); document.write( "Thank you
\n" ); document.write( "

Algebra.Com's Answer #46171 by 303795(602)\"\" \"About 
You can put this solution on YOUR website!
If you have a total of 400 feet of timber then you cannot have a patio length of 400 feet.\r
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\n" ); document.write( "\n" ); document.write( "2L = 400 - 2W <-- so L=200-W
\n" ); document.write( "A=LW
\n" ); document.write( "=(400-2W)W <-- =(200-W)W
\n" ); document.write( "=4==W-2W^ <-- =200W - W^2
\n" ); document.write( "A= 2W^ - 400W <-- Area = -W^2+200W\r
\n" ); document.write( "\n" ); document.write( "Not sure what you mean by vertex form but the line of symmetry is at (-b/2a)
\n" ); document.write( "=-200/-2
\n" ); document.write( "=100
\n" ); document.write( "So a width of 100 and a length of 100 will use 400 feet of timber and have an area of 10000 square feet.\r
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