2x-y + 2 ≥ 0, X-1 ≤ 0, 0 ≥ y ≥ 2, z = 3x_ The minimum value of 2Y

2x-y + 2 ≥ 0, X-1 ≤ 0, 0 ≥ y ≥ 2, z = 3x_ The minimum value of 2Y




What is the minimum value of y = (x ^ 2 + 1) ^ 2 / [(3x ^ 2 + 2) (2x ^ 2 + 3)]~


y=(x^4+2x^2+1)/(6x^4+13x^2+6)
=(x^2+x^(-2)+2)/(6x^2+6x^(-2)+13)
Let x ^ 2 + x ^ (- 2) + 2 = K
Then y = K / (6K + 1) = 1 / (6 + 1 / k)
So as long as K is the minimum
Because x ^ 2 + x ^ (- 2) > = 2
So Kmin = 4
So Ymin = 4 / 25

Elden Ring Premiere

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