Find the maximum and minimum values of the quadratic form subject to the constra

Question

Find the maximum and minimum values of the quadratic form subject to the constraint x12+x22+x32=1 , and determine the values of x1 ,x2 ,x3 and at which the maximum and minimum occur.

x12+x22+4x32-2x1x2+8x1x3+8x2x3

Give exact answers.

Explanation / Answer

x1^2 + x2^2 = 1 - x3^2 ( from constraint) so expression becomes =1 -x3^2 +4x3^2 -2x1x2 +8x1x3 +8x2x3 =1 +3x3^2 -2x1x2 +8x1x3 +8x2x3 note that (x1+x2)^2 = x1^2 +x2^2 +2x1x2 so -2x1x2 = (x1^2 + x2^2)-(x1 + x2)^2 = 1-x3^2 - (x1+x2)^2 so expression becomes 1 + 3x3^2 +1-x3^2 - (x1+x2)^2+8x1x3+8x2x3 = 2 +2x3^2+ - (x1+x2)^2+ 8x3( x1+x2) now let y = (x1+x2) 2+ 2x3^2+ - y^2+ 8x3y diff w.r.t. y, set to 0 -2y + 8x3 = 0 => y = 4x3 diff wrt x3,. set to 0 4x3+8y = 0 => y = (-1/2)x3 First Lets look at y = 4x3 Then expression becomes =2 +2x3^2+ - (4x3)^2+ 8x3( 4x3) =2+ 10x3^2 Clearly smallest value of expression is when x3=0, at which expression = 2, But if x3=0, y = x1+x2 = 0, so x1=-x2 But x1^2 + x2^2 = 1-x3^2, which works out to x1=1/sqrt(2), x2 = -1/sqrt(2), x3=0, expression=2 Now lets look at y=(-1/2)x3 Then expression becomes =2 +2x3^2+ - (-1/2)x3)^2+ 8x3( (-1/2)x3) =2- (9/4)(x3^2) Clearly largest value of expression is when x3=0, at which expression = 2, But if x3=0, y = x1+x2 = 0, so x1=-x2 But x1^2 + x2^2 = 1-x3^2, which works out to x1=1/sqrt(2), x2 = -1/sqrt(2), x3=0, expression=2 So the minima & maxima occur at same values of x1,x2,x3.

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