Suppose we have the following identity:
(px + (1-p)y)^2 = Ax^2 + Bxy + Cy^2.
Find the minimum of max(A,B,C) over 0 \leq p \leq 1.
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To find the minimum of max(A, B, C) over 0 ≤ p ≤ 1, we need to analyze the equation (px + (1 - p)y)^2 = Ax^2 + Bxy + Cy^2.
Let's start by expanding the left side of the equation:
(px + (1 - p)y)^2 = (px)^2 + 2px(1 - p)y + ((1 - p)y)^2
= p^2x^2 + 2pxy - 2p^2xy + (1 - 2p + p^2)y^2
= (p^2 - 2p + 1)x^2 + (2p - 2p^2)xy + (1 - 2p + p^2)y^2.
By comparing this expanded form with the right side of the equation, Ax^2 + Bxy + Cy^2, we can conclude that:
A = p^2 - 2p + 1
B = 2p - 2p^2
C = 1 - 2p + p^2.
To find the minimum of max(A, B, C), we need to analyze the behavior of A, B, and C individually as p varies from 0 to 1.
Let's analyze each term:
1. A = p^2 - 2p + 1:
To find the minimum of A, we can take the derivative with respect to p and set it equal to zero:
dA/dp = 2p - 2 = 0
2p = 2
p = 1.
Since p = 1 lies within the range 0 ≤ p ≤ 1, we need to evaluate A at both endpoints and p = 1:
A(0) = (0)^2 - 2(0) + 1 = 1
A(1) = (1)^2 - 2(1) + 1 = 0.
2. B = 2p - 2p^2:
To find the minimum of B, we can take the derivative with respect to p and set it equal to zero:
dB/dp = 2 - 4p = 0
4p = 2
p = 1/2.
Since p = 1/2 lies within the range 0 ≤ p ≤ 1, we need to evaluate B at both endpoints and p = 1/2:
B(0) = 2(0) - 2(0)^2 = 0
B(1) = 2(1) - 2(1)^2 = 0
B(1/2) = 2(1/2) - 2(1/2)^2 = 1/2.
3. C = 1 - 2p + p^2:
To find the minimum of C, we can take the derivative with respect to p and set it equal to zero:
dC/dp = -2 + 2p = 0
2p = 2
p = 1.
Since p = 1 lies within the range 0 ≤ p ≤ 1, we need to evaluate C at both endpoints and p = 1:
C(0) = 1 - 2(0) + (0)^2 = 1
C(1) = 1 - 2(1) + (1)^2 = 0.
Finally, we compare the values of A, B, and C at the endpoints and critical points:
Minimum of max(A, B, C) = min(max(1, 0, 1/2), max(1, 0, 1), max(1, 0, 0))
= min(1, 1, 1)
= 1.
Therefore, the minimum value of max(A, B, C) over the interval 0 ≤ p ≤ 1 is 1.
To find the minimum of max(A, B, C) over the given range of p, we need to compute the values of A, B, and C for any given value of p and then find the minimum among the maximum values.
Let's begin by expanding the square on the left-hand side of the given identity:
(px + (1 - p)y)^2 = (px)^2 + 2px(1 - p)y + ((1 - p)y)^2
Simplifying the above expression, we get:
(px + (1 - p)y)^2 = p^2x^2 + 2pxy - 2p^2xy + (1 - p)^2y^2
Expanding further, we have:
(px + (1 - p)y)^2 = p^2x^2 + 2pxy - 2p^2xy + y^2 - 2(1 - p)xy + (1 - p)^2y^2
Comparing this to the expression Ax^2 + Bxy + Cy^2, we can equate the corresponding coefficients:
A = p^2, B = -2p^2 + 2(1 - p), C = (1 - p)^2
To find the minimum of max(A, B, C), we need to find the minimum of the maximum value among A, B, and C for any given p.
1. A = p^2: The maximum value of A occurs at p = 1, where A = 1^2 = 1.
2. B = -2p^2 + 2(1 - p):
To find the maximum value of B, we can take the derivative with respect to p and equate it to 0:
d(B)/dp = -4p + 2 = 0
Solving for p, we get p = 1/2.
Plugging p = 1/2 into B, we get B = -2(1/2)^2 + 2(1 - (1/2)) = -1/2.
Therefore, the maximum value of B is -1/2.
3. C = (1 - p)^2: The maximum value of C occurs at p = 0, where C = (1 - 0)^2 = 1.
Comparing the maximum values of A, B, and C, we find that the minimum of max(A, B, C) is -1/2.
Therefore, the minimum value of max(A, B, C) over 0 ≤ p ≤ 1 is -1/2.