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 #772210 by Edwin McCravy(20060)\"\" \"About 
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document.write( "I consider this as a 3D problem in Calc 3.\r\n" );
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document.write( "The second partials test:\r\n" );
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document.write( "If (a,b) is a point on f(x,y) for which both partial derivatives are 0, then\r\n" );
document.write( "find the quantity\r\n" );
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document.write( "d = fxx(a,b)∙fyy(a,b)-[fxy(a,b)]2\r\n" );
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document.write( "1. If d is negative, then (a,b) is a saddle point which is neither a relative\r\n" );
document.write( "   maximum nor minimum.\r\n" );
document.write( "2. If d is positive\r\n" );
document.write( "   a. if fxx(a,b) > 0, then (a,b) is a relative minimum\r\n" );
document.write( "   b. if fxx(a,b) < 0, then (a,b) is a relative maximum\r\n" );
document.write( "3. If d is 0, the test fails.\r\n" );
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document.write( "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)?\r\n" );
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document.write( "f(x,y) = p(x+y)-q(x+y) = (2x^2 +3y^2)-(4x-18y-39) = 2x²+3y²-4x+18y+39,\r\n" );
document.write( "which is an elliptical paraboloid whose axis of symmetry is parallel to the \r\n" );
document.write( "z-axis, and has just one minimum point or one maximum point.\r\n" );
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document.write( "We set the two partial derivatives of f equal to 0:\r\n" );
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document.write( "fx=px-q=4x-4, fy=py-qy=6y+18\r\n" );
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document.write( "     4x-4=0,   6y+18=0\r\n" );
document.write( "        x=1,       y=-3\r\n" );
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document.write( "So if that's the minimum point, that's the one we use to evaluate the minimum\r\n" );
document.write( "value.  But we ought to show that it's a minimum\r\n" );
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document.write( "We calculate d\r\n" );
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document.write( "d = fxx(1,-3)∙fyy(1,-3)-[fxy(1,-3)]2\r\n" );
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document.write( "fxx(1,-3) = 4, fyy(1,-3) = 6, ffxx(1,-3) = 0, \r\n" );
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document.write( "d = 4∙6-0² = 24 > 0}\r\n" );
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document.write( "So since d = 24 > 0 and fxx(1,-3) = 4 then f does have a relative minimum\r\n" );
document.write( "at (1,-3).\r\n" );
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document.write( "That minimum value is found by substituting\r\n" );
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document.write( "f(1,-3) = p(1+(-3))-q(1+(-3)) = 2(1)²+3(-3)²-4(1)+18(-3)+39,\r\n" );
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document.write( "That works out to be 10.\r\n" );
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document.write( "Answer: 10\r\n" );
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document.write( "Edwin
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