document.write( "Question 1103715: A piece of wire 64 cm long is cut into two pieces with different lengths. The two pieces are formed to TWO squares.
\n" ); document.write( " (a) Show that the combined area of the two squares A is given by A=2x^2-30x+256
\n" ); document.write( " (b) What is the minimum combined area of the two squares?
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Algebra.Com's Answer #718437 by math_helper(2461)\"\" \"About 
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A piece of wire 64 cm long is cut into two pieces with different lengths. The two pieces are formed to TWO squares.
\n" ); document.write( " (a) Show that the combined area of the two squares A is given by A=2x^2-30x+256
\n" ); document.write( " (b) What is the minimum combined area of the two squares?
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\n" ); document.write( "\n" ); document.write( "Let x = the length of the first piece
\n" ); document.write( "the 64-x = the length of the 2nd piece\r
\n" ); document.write( "\n" ); document.write( "Area of the first square is \"+%28x%2F4%29%5E2+\"
\n" ); document.write( "Aree of the 2nd square is \"+%28%2864-x%29%2F4%29%5E2+\"\r
\n" ); document.write( "\n" ); document.write( "Combined area = \"+%28x%5E2%2F16%29+%2B+%28%284096-128x%2Bx%5E2%29%2F16%29+\"
\n" ); document.write( "This simplifies to \"+highlight%28matrix%281%2C3%2C+%22+%22%2C+%281%2F8%29%28x%5E2-64x%2B2048%29%2C%22+%22%29%29+\"\r
\n" ); document.write( "\n" ); document.write( "Notice that this disagrees with the value of A in the problem statement. I checked my work but I've been known to make errors.
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\n" ); document.write( "To find the minimum, take the derivative. I will use my equation (the process is the same even if the correct equation is different):
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\n" ); document.write( " \"+A+=+%281%2F8%29%28x%5E2-64x%2B2048%29+\"
\n" ); document.write( "Take the derivative of A with respect to x:
\n" ); document.write( " \"+dA%2Fdx+=+%281%2F8%29%282x+-+64%29+\"
\n" ); document.write( "Set it to zero:
\n" ); document.write( " \"+++%281%2F8%29%282x-64%29+=+0++\" —>
\n" ); document.write( "\"x+=+32cm\" \r
\n" ); document.write( "\n" ); document.write( " Min Area = \"+%2832%2F4%29%5E2+%2B+%2832%2F4%29%5E2+=+8%5E2+%2B+8%5E2+=+highlight%28128cm%5E2%29+\" \r
\n" ); document.write( "\n" ); document.write( "This is a local minimum of A because \"+d%5E2A%2Fdx%5E2+=+%281%2F8%29%282%29+\" —> concave up. Normally, you should also check the endpoints (here x=0, x=64) but due to the quadratic form, we know the local minimum is a global minimum on the interval.\r
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