Given that real numbers x and y satisfy the conditions y ≤ 0, y ≥ x, 2x + y + 4 ≥ 0, then the minimum value of Z = x + 3Y is

Given that real numbers x and y satisfy the conditions y ≤ 0, y ≥ x, 2x + y + 4 ≥ 0, then the minimum value of Z = x + 3Y is

This is a linear programming problem
The feasible regions are as follows
Objective function, z = x + 3Y
∴ y=(-1/3)x+z/3
It's a set of lines with a slope of - 1 / 3
When a straight line passes through B, the longitudinal intercept is the minimum, and Z has the minimum value
The optimal solution is (- 4 / 3, - 8 / 3)
The minimum value of Z is - 4 / 3 + 3 * (- 8 / 3) = - 28 / 3

If real numbers x and y satisfy (x ^ 2 / 16) + (y ^ 2 / 9) = 1, then z = the maximum and minimum of X-Y?

Let x = 4sina, y = 3cosa
z=x-y=4sina-3cosa=5sin(a-b)
SINB = 3 / 5, CoSb = 4 / 5
-5≤z=5sin(a-b)≤5
The maximum and minimum are 5 and - 5
[questions are welcome]

The maximum and minimum values of x ^ 2 / 4 + y ^ 2 / 2 = 1, x ^ 2 + y ^ 2-x are known

X ^ 2 / 4 + y ^ 2 / 2 = 1 Let x = 2cosa y = √ 2sinax ^ 2 + y ^ 2-x = 4cos ^ 2A + 2Sin ^ 2a-2cosa = 4cos ^ 2A + 2-2cos ^ 2a-2cosa = 2cos ^ 2a-2cosa + 2 = 2 (cosa-1 / 2) ^ 2 + 3 / 2, because - 1 ≤ Cosa ≤ 1, the minimum value in 2 (cosa-1 / 2) ^ 2 + 3 / 2 is 3 / 2, and the maximum value is 6