Given

subject to the constraint

Let

.
The gradient vectors of

and

are:

and

By Lagrange's theorem, there is a number

, such that


It can be seen that

has local extreme values at the given region.
Answer:
y=-2/3x+20/3
Step-by-step explanation:
m=(y2-y1)/(x2-x1)
m=(4-6)/(4-1)
m=-2/3
y-y1=m(x-x1)
y-6=-2/3(x-1)
y=-2/3x+2/3+6
y=-2/3x+2/3+18/3
y=-2/3x+20/3
Answer:
k = 15.94
Step-by-step explanation:
Answer: AB=12
Los triangulos BCD y ADE son iguales por tener dos lados y el angulo comprendido respectivamente iguales. por tanto AE=DC=4
Luego AB = AE+EB = 4+8 = 12
Answer:
1, 7, 17, 31
Step-by-step explanation:
2(1)² + 1 - 2 = 3-2 = 1
2(2)² + 1 - 2 = 9-2 = 7
2(3)² + 1 - 2 = 19-2 = 17
2(4)² + 1 - 2 = 33-2 = 31