Convex Optimization. Mikhail Moklyachuk
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Название: Convex Optimization

Автор: Mikhail Moklyachuk

Издательство: John Wiley & Sons Limited

Жанр: Математика

Серия:

isbn: 9781119804086

isbn:

СКАЧАТЬ It remains to be assumed that λ1 > 0. Then image. Given that λ1 ≠ 0, λ2 ≠ 0, we deduce from this that image, and therefore λ2 = 0. Now we can easily get another group of solutions of the system:

image

      Note that in (1) and (3) the multiplier λ0 can accept both positive and negative values, in (2) and (4) it can accept only positive values and in (5) it can accept only negative values. Therefore, (0, 1) and (1, 0) are stationary points both in the minimization problem and in the maximization problem, (0, –1) and (–1, 0) are stationary points only in the minimization problem, and the point of (5) is the stationary point only in the problem of maximization.

      Now we will study the stationary points for optimality. The function f is strongly convex on ℝ2. Therefore, it reaches the global minimum on any closed set X. We calculate the value of f at the stationary points of the minimization problem:

image

      Since a < b, hence (1.0) and (–1.0) are points of the global minimum of the function f on X.

      Let us represent the function f in the form

image

      If we move from the points (0, 1) and (0, –1), remaining on the circle image, and hence in X, then the value of f will decrease. Consequently, these points are not points of the local minimum f on X. At the same time, for any ε > 0, the point (–ε, 1) belongs to X and f(0, 1) < f(–ε, 1). Therefore, the point (0, 1) is not a point of the local maximum f on X. Consequently, the stationary points (0, 1) and (0, –1) are not solutions of the problem.

      We now consider the matrix of the second derivatives of the Lagrangian function:

image

      For values with (5), this matrix is as follows:

image

      Since λ0 < 0, this matrix is positive definite. Sufficient conditions for the minimum are fulfilled. Consequently, (image, image) is the point of the strict local minimum of f on X.

      Answer. image ∈ absmin, image ∈ absmin, image ∈ absmax. Δ

      Let us solve the following optimization problems.

      1 1) f(x, y) = x4 + y4 − 4xy → extr.

      2 2) f(x, y) = ae−x + be−y + ln(ex + ey) → extr.

      3 3) f(x, y) = (x + y)(x − a)(y − b) → extr.

      4 4) f(x, y) = x2 − 2xy2 + y4 − y5 → extr.

      5 5) f(x, y) = x + y + 4 sin (x) sin (y) → extr.

      6 6) f(x, y) = xex − (1 + ex) cos (y) → extr.

      7 7) f(x, y) = (x2 + y2)e−(x2 + y2) → extr.

      8 8) f(x, y) = xy ln (x2 + y2) → extr.

      9 9) .

      10 10) f(x, y) = sin (x) sin (y) sin(x + y) → extr, 0 ≤ x ≤ π, 0 ≤ y ≤ π.

      11 11) f(x, y) = sin (x) +cos (y) +cos (x − y) → extr, 0 ≤ x ≤ π/2, 0 ≤ y ≤ π/2.

      12 12) f(x, y) = x2 + xy + y2 − 4 ln (x) − 10 ln (y) → extr.

      13 13) f(x, y) = (5x + 7y − 25)e−(x2 + y2 + xy) → extr.

      14 14) f(x, y) = ex2−y (5 − 2x + y) → extr.

      15 15) f(x, y) = e2x+3y(8x2 − 6xy + 3y2) → extr.

      16 16) .

      17 17) .

      18 18) .

      19 19) f(x, y) = 2x4 + y4 − x2 − 2y2 → extr.

      20 20) f(x, y) = x2 − xy + y2 − 2x + y → extr.

      21 21) f(x, y) = xy + 50/x + 20/y → extr.

      22 22) f(x, y) = x2 − y2 − 4x + 6y → extr.

      23 23) f(x, y) = 5x2 + 4xy + y2 − 16x − 12y → extr.

      24 24) f(x, y) = 3x2 + 4xy + y2 − 8x − 12y → extr.

      25 25) f(x, y) = 3xy − x2y − xy2 → extr.

      26 26) f(x, y, z) = x2 + y2 + z2 − xy + x − 2z → extr.

      27 27) f(x, y, z) = x2 + 2y2 + 5z2 − 2xy − 4yz − 2z → extr.

      28 28) f(x, y, z) = xy2z3(a − x − 2y − 3z) → extr, a > 0.

      29 29) f(x, y, z) = x3 + y2 + z2 + 12xy + 2z → extr, x > 0, y > 0, z > 0.

      30 30) f(x, y, z) = x + y2/4x + z2/y + 2/z → extr.

      31 31) f(x, y, z) = x2 + y2 + z2 + 2x + 4y − 6z → extr.

      32 32) f(x, y) = y → extr, x3 + y3 − 3xy = 0.

      33 33) f(x, y) = x3 + y3 → extr, ax + by = 1, a > 0, b > 0.

      34 34) f(x, y) = x3/3 + y → extr, x2 + y2 = a, a > 0.

      35 35) f(x, y) = x sin (y) → extr, 3x2 − 4 cos (y) = 1.

      36 36) f(x, y) = x/a + y/b → extr, x2 + y2 = 1.

      37 37) f(x, y) = x2 + y2 → extr, x/a + y/b = 1.

      38 38) f(x, y) = Ax2 + 2Bxy + Cy2 → extr, x2 + y2 = 1.

      39 39) f(x, y) = x2 + 12xy + 2y2 → extr, 4x2 + y2 = 25.

      40 40) f(x, y) = cos2 (x) + cos2 (y) → extr, x − y = π/4.

      41 41) f(x, y) = x/2 + y/3 → extr, x2 + y2 = 1.

      42 42) f(x, y) = x2 + y2 → extr, 3x + 4y = 1.

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