document.write( "Question 1150766: if p(x+y) = 2x² +3y², and q(x+y)=4x-18y-39, where x and y are real numbers, what is the minimum value of p(x+y)-q(x+y)? \n" ); document.write( "

Algebra.Com's Answer #772197 by ikleyn(52834) $\"\"$ $\"About$

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document.write( "The difference is\r\n" );
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document.write( "    p(x+y)-q(x+y) = 2x^2 + 3y^2 - 4x + 18y + 39\r\n" );
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document.write( "Continue by completing the squares\r\n" );
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document.write( "    = (2x^2 - 4x)      + (3y^2 + 18y) + 39\r\n" );
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document.write( "    = 2*(x^2 - 2x)     + 3*(y^2 + 6y) + 39 =\r\n" );
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document.write( "    = 2*(x^2 - 2x + 1) + 3*(y^2 + 6y + 9) + 39 - 2 - 27 =\r\n" );
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document.write( "    = 2*(x-1)^2        + 3*(y+3)^2        + 10.\r\n" );
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document.write( "The minimum is achieved at x= 1,  y= -3  and is equal to  10.    ANSWER\r\n" );
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