When x and y are the values, the square of x plus the square of Y minus 4x plus 6y plus 18 has the minimum value, and this minimum value can be obtained

When x and y are the values, the square of x plus the square of Y minus 4x plus 6y plus 18 has the minimum value, and this minimum value can be obtained

x^2+y^2-4x+6y+18
=(x²-4x+4)+(y²+6y+9)+5
=(x-2)²+(y+3)²+5
So x = 2, y = - 3
Minimum = 5

The maximum value of function y = (x-1) ^ 2 + 4 on [- 1,5] is 20, and the minimum value is? Specific process, thank you

Function y = (x-1) ^ 2 + 4
Axis of symmetry x = 1
So x = 1, minimum = 4
x-1 y=8
x=5 y=20
Max = 20

If x, y satisfy x + y ≥ 1, X-Y ≥ - 1, 2x-y ≤ 2. The objective function z = ax + 2Y only obtains the minimum value at (1,0), then the value range of a is obtained
-A / 2 x + Z / 2 = y, the slope of this equation is - 1

Let the variables X and y satisfy the constraint condition x + y ≥ 1 x − y ≥ 12 x − y ≤ 2. The objective function z = ax + 2 y only obtains the minimum value at (1,0), then the value range of a is ()
A. （-1，2）B. （-2，4）C. （-4，0]D. （-4，2）

When a ＞ 0, the slope of the line ax + 2y-z = 0 K = - A2 ＞ KAC = - 1, the solution is a ＜ 2. When a ＜ 0, k = - A2 ＜ KAB = 2, the solution is a ＞ - 4