To find the extremum of the function z = x2-xy + y2-2x + y

To find the extremum of the function z = x2-xy + y2-2x + y

I will use DZ / DX to express the partial derivative of Z to x, d ^ 2Z / DXDY to express the partial derivative of Z to x, y to express the partial derivative of Z to x, y to express the partial derivative of Z to x, d ^ 2Z / DX ^ 2 to express the second order partial derivative of Z to X. DZ / DX = 2x-y-2, DZ / dy = 2y-x + 1D ^ 2Z / DX ^ 2 = 2, d ^ 2Z / dy ^ 2 = 2, d ^ 2Z / D

The extremum of F (x, y) = x ^ 2 + XY + y ^ 2-3x-6y

I don't know how old you are?
In college mathematics, there is a way to seek complete derivation;
F (x, y) first differentiates X and treats y as a constant. Then FX '= 2x + Y-3
Then we take x as a constant and derive y. then FY '= x + 2y-6
Only when the two formulas are equal to 0, is to go to the extremum, as for the maximum or minimum, or does not exist, we do not consider here
We can know that there is only one extremum when x = 0 and y = 3,
f(0,3)=-9
Let's go in any number, for example, f (x, y) = 0 from (0,0)
So the extremum is a minimum

Finding the extreme value of function f (x, y) = x ^ 2 + XY + y ^ 2-3x-6y

The derivation of X and Y in F (x, y) shows that f '(x) = 2x + Y-3, f' (y) = x + 2y-6
Let f '(x) = 0, f' (y) = 0, the roots of the two equations are (0,3)
Then find out the answer with the root!

Stationary point of binary function f (x, y) = x ^ 2 + XY + y ^ 2-3x-6y

fx=2x+y-3=0
fy=x+2y-6=0
x=0,y=3
(0,3)

If the extremum function bx2 + = - 3 is known, then F-X is taken as X-1
1. Discuss whether f (1) and f (- 1) are the maximum or minimum of F (x)
2. The tangent of the curve y = f (x) is made through a (0,16), and the tangent equation is solved

Derivation of F (x)
(1)
f `（x）=3ax^2+2bx-3
Given the function f (x) = ax ^ 3 + BX ^ 2-3x, take the extremum f '(1) = 0 f' (- 1) = 0 at x = + - 1
The solution is a = 1, B = 0
f ``(x)=6x f ``(1)=6>0 f ``(-1)

Finding the extremum of function f (x, y) = XY + 1 / x + 1 / Y

Finding partial derivative of x = Y-1 / x ^ 2 = 0
Finding partial derivative of y = X-1 / y ^ 2 = 0
Get x = 1, y = 1
So the extremum is 3
Find a second-order partial derivative, you can see is a minimum

What is the extremum of the function f (x, y) = XY (x + Y-9)

The necessary condition for the function to obtain the extremum is that the partial derivative of F to X is zero, and the partial derivative of F to y is zero
Y (x + Y-9) + xy = 0; X (x + Y-9) + xy = 0. By solving the equations, we get x = 3, y = 3, that is, the extreme point of the function is (3,3)

If the three planes are perpendicular to each other, their three intersecting lines intersect at a point O, and the distances from P to the three planes are 3, 4 and 5 respectively, then the length of OP is______ ．

Construct cuboids with edge lengths of 3, 4 and 5, so that OP is the diagonal of the cuboid. So OP = 32 + 42 + 52 = 52

Verification: three planes intersect each other to get three intersection lines. If two of them intersect at a point, then the third one also passes through this point

It is proved that let three planes be α, β, γ,
And α ∩ β = C, α ∩ γ = B, β ∩ γ = a; ∩ α ∩ β = C, α ∩ γ = B, ∩ C ∩ 8834; α, B ∩ 8834; α;
C and B intersect at a point or are parallel to each other
(1) As shown in Figure 1, if C and B intersect at a point, let C ∩ B = P
From P ∈ C, and C &; β, there is p ∈ β; from P ∈ B, B &; γ, there is p ∈ γ; ∩ γ = a; from P ∈ B, B &; γ, there is p ∈ γ; ∩ P ∈;
Therefore, the line a, B and C intersect at a point (P)
Figure 1; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; Figure 2
(2) As shown in Figure 2, if C ∥ B, then B ∥ 8834; γ, and C ∥ 8836; γ, ∥ C ∥ γ; and C ∥ 8834; β,
So a, B and C are parallel to each other

Prove that three planes intersect each other. If the three intersecting lines are a, B and C, then ABC intersects at one point

If a is parallel to B, i.e. a is parallel to a, then plane B made through a intersects plane a at C, so a ‖ C when a is not, a and B intersect at a point D, D is on plane a and B, so it is on the intersection line of AB, so D is on C, so ABC intersects at a point