Answer:
Point Critical point
Q (2,0) local minimum
R (-2,1) Saddle
S (2,-1) local maximum
T ( -2,-1) Saddle
O ( -2,0) Saddle
Step-by-step explanation: INCOMPLETE ANSWER INFORMATION ABOUT THE POINTS ARE RARE
f(x,y) = x³ +y⁴ - 6x -2y² +3
df/dx = f´(x) = 3x² -6x
df/dxdx = f´´(xx) = 6x
df/dy = f´(y) = -4y
df/dydy = 4
df/dydx = df/dxdy = 0
df/dydy = f´´(yy)
D = [ df/dxdx *df/dydy] - [df/dydx]²
D = (6x)*4 - 0
D = 6*2*4 D > 0 and the second derivative on x is 6*2 = 12
so D > 0 and df/dxdx >0 there is a local minimum in P
Q(2,1)
D = (6*2)*4 D>0 and second derivative on x is 6*2
The same condition there is a minimum in Q
R ( -2,1)
D = 6*(-2)*4 = -48 D< 0 there is a saddle point in R
S (2,-1)
D = 6*2*4 = 48 D > 0 and df/dxdx = 6*-1 = -6
There is a maximum in S
T ( -2,-1)
D = 6*(-2)*(4) = -48 D<0 there is a saddle point in T
O ( -2,0)
D = 6*(-2)*4 = -48 D<0 there is a saddle point in O
a) The vertices A, B, C of a triangle are (2, -1, -3), (4, 2, 3) and (6, 3, 4) respectively. Show that AB = (2,3,6) and AC = 9.
Sav [38]
Answer:
... a triangle are 2,-1,-3 4,2,3 and 6,3,4 respectively. If vector AB=2,3,6 vector AC=4,4,7 vector BC=2,1,1 , show that AB=7 AC=9 and BC= square root of 6.
Answer:
1) x- y = 8
x + y = 12
eliminate one variable by adding the equations
2x = 20
divided by 2 on both sides
x = 10
substitute the value of x in the equation 1
10 - y = 8
-y = 8- 10
y =2
[x , y]
[10 , 2]
2) 2x - y = 4
3x + y = 6
eliminate one variable by adding the equations
5x = 10
x = 2
substitute the value of x in the equation
2(2) - y = 4
4 - y = 4
y = 0
[x ; y]
[2; 0]
3) x + 2y = 10
x + 4y = 14
multiply equation 1 by - 1
-x -2y = -10
x + 4y = 14
eliminate one variable by adding the equations
2 y = 4
y = 2
substitute the value of y in the equation 1
x + 2(2) = 10
x + 4 = 10
x = 6
[x ; y]
[6 ; 2]
4 ) 3x + y = 9
y = 3x + 6
simplify the expression
3x + y = 9
- 3x + y = 6
eliminate one variable by adding the equations
2y = 15
y = 15/2
substitute the value of y in equation 1
3x + 15/ 2 = 9
3x = 3/2
x = 1/2
[x ; y]
[1/2 ; 15/ 2]
5) 4x + 5y = 15
6x - 5y = 18
eliminate one variable by adding the equations
10 x = 33
× = 33/10
substitute the value of x in equation I
4 (33/10) + 5y = 15
66/5 + 5y =15
5y = 9/5
y = 9/25
[x ; y]
[33/10 ; 9/25]
6) 5x = 7y
x + 7y = 21
in equation I move the variable to the left
5x - 7y = 0
x + 7 y = 21
eliminate one variable by adding the equations
6x = 21
x = 7/2
substitute the value of x in the equation
5(7/2) = 7y
35/2 = 7y
5/2 = y
[x ; y ]
[7/2 ; 5/2 ]
No, because the sum of the total angle measure (or whatever it is called) is 720°, but those numbers only add up to 710°
