Let
f ( x , y ) = x 2 + 2 x y + 3 y 2
where x , y ∈ R satisfy,
4 x 2 + 2 x y + 9 y 2 = 1 0 0
Find the minimum of f ( x , y ) . The minimum can be expressed as q p where p , q are positive coprime integers. As your answer, enter p + q
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We first find whether there's any critical points, c inside the domain D = { ( x , y ) ∣ 4 x 2 + 2 x y + 9 y 2 = 1 0 0 } , this is equivalent to finding ∇ f ( c ) = 0 (since ∇ f ( c ) must exist as f is a multivariate polynomial), so ∇ f = ⟨ 2 x + 2 y , 6 y + 2 x ⟩ = ⟨ 0 , 0 ⟩ which means c = ⟨ x , y ⟩ = ⟨ 0 , 0 ⟩ but it does satisfy the given constraint.
Now, we compute the extrema of f on ∂ D , the boundary of D , using the method of Lagrange Multipliers, let g ( x , y ) = 4 x 2 + 2 x y + 9 y 2 − 1 0 0 = 0 as our constraint, we have ∇ f = λ ∇ g ⟨ 2 x + 2 y , 6 y + 2 x ⟩ = λ ⟨ 8 x + 2 y , 2 x + 1 8 y ⟩ . It is obvious that ∇ g = 0 , otherwise we have ( x , y ) = ( 0 , 0 ) which doesn't satisfy our constraint. Also, if λ = 0 , then we also get ( x , y ) = ( 0 , 0 ) which also doesn't satisfy our constraint.
Then, eliminating λ we have ( 2 x + 2 y ) ( 1 8 y + 2 x ) = ( 8 x + 2 y ) ( 6 y + 2 x ) ⟹ x 2 + x y = 2 y 2 ⟹ ( 2 x + y ) 2 = 9 y 2 ⟹ x = y or x = − 2 y . If x = y , then from g , we have 1 5 x 2 = 1 0 0 , x 2 = 3 2 0 , so f = 6 x 2 = 4 0 . If x = − 2 y , then from g , we have 2 1 y 2 = 1 0 0 , y 2 = 2 1 1 0 0 , so f = 3 y 2 = 7 1 0 0 .
Thus, f m a x = 4 0 and f m i n = 7 1 0 0 , p + q = 1 0 0 + 7 = 1 0 7 .
I did exactly the same. Very nice.
If we define the symmetric matrices A = ( 1 1 1 3 ) B = ( 4 1 1 9 ) then we want to minimize f ( x ) = x T A x subject to the condition g ( x ) = x T B x = 1 0 0 . Find an orthogonal matrix P and real numbers u , v such that B = P T ( u 0 0 v ) P Since g ( x , y ) = 3 x 2 + ( x + y ) 2 + 8 y 2 , it is clear that g ( x , y ) is positive definite and hence u , v > 0 . Thus, if we define Q = ( u 0 0 v ) P then Q is a nonsingular matrix and B = Q T Q . If we change coordinates, and define X = Q x , then we are trying to minimize X T ( Q − 1 ) T A Q − 1 X subject to X T X = 1 0 0 Since ( Q − 1 ) T A Q − 1 is symmetric, we can find an orthogonal matrix S and reals a < b such that S T ( Q − 1 ) T A Q − 1 S = ( a 0 0 b ) Changing coodinates once again, defining ( ξ 1 ξ 1 ) = S T X , we end up trying to minimize a ξ 1 2 + b ξ 2 2 subject to the restraint ξ 1 2 + ξ 2 2 = 1 0 0 , and so the minimum value is 1 0 0 a , and the maximum value is 1 0 0 b . Now ( t − a ) ( t − b ) = d e t ( t I 2 − S T ( Q − 1 ) T A Q − 1 S ) = d e t ( S T ( t I 2 − ( Q − 1 ) T A Q − 1 ) S ) = d e t ( t I 2 − ( Q − 1 ) T A Q − 1 ) = d e t ( ( Q − 1 ) T ( t Q T Q − A ) Q − 1 ) = ( d e t Q ) − 2 d e t ( t B − A ) and hence we need to solve 0 = d e t ( t B − A ) = 3 5 t 2 − 1 9 t + 2 = ( 5 t − 2 ) ( 7 t − 1 ) and hence a = 7 1 , while b = 5 2 . Thus the minimum value is 7 1 0 0 , making the answer 1 0 7 , and the maximum value is 4 0 .
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Let the value of f(x,y)=c for point (a,b) such that this point also lies on the second curve. Thus we have 2 equations a²+2ab3b²=c and 4a²+2ab+9b²=100. Since we need to find the max/min value of c, we can imagine in either case both curves must touch each other at the point (a,b)[because the point must be unique and no other point must satisfy the condition within the close neighborhood of the point, otherwise it would be a case of cutting and max/min won't be achieved]. This means that the slopes of both curves must be equal at (a,b). That gives us another condition (a+b)/(a+3b)=(4a+b)/(a+9b) [via elementary calculus, just differentiate the curves and equate dy/dx of both curves], which on simplifying turns out to be a²+ab-2b²=0 =>(a-b)(a+2b)=0 => a=b or a=-2b. Now out of the 2 cases, one is for maximum and other for the minimum value.
Case 1, a=b, we have 15b²=100, 6b²=c => c=40 [by eliminating b²].
Case 2, a=-2b, we have 3b²=c, 21b²=100 => c=100/7.
Thus the min value is 100/7.