It remains to be assumed that λ1 > 0. Then 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:
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
If we move from the points (0, 1) and (0, –1), remaining on the circle
We now consider the matrix of the second derivatives of the Lagrangian function:
For values with (5), this matrix is as follows:
Since λ0 < 0, this matrix is positive definite. Sufficient conditions for the minimum are fulfilled. Consequently, (
Answer.
1.5. Exercises
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.
