X+2y≤6 with x≥0 and y≥0
2x+y≤6
1st) Find the 1st shaded region: 1st Quadrant since x≥0 and y≥0
2nd) Let's find the x and y intercepts of 2x+y = 6
x-intercepts for y = 0 , x=6 →A(3,0)
y-intercepts for x = 0 , y =3→B(0,6) . Now Join AB
2nd shaded region
3rd) Let's find the x and y intercepts of x+2y = 6
x-intercepts for y = 0 , x=6 →C(6,0)
y-intercepts for x = 0 , y =3→D(0,3) . Now Join CD
3rd shaded region
4th) Now let's calculate the coordinate of the intersection point of
x+2y=6 and
2x+y =6 Solving it will give you x = 2 and y = 2 (coordinates of the intersection of AB with DC, le be E. (4th shaded region).
Re write all pairs:
A(3,6), B(0,6), C(6,0), D(0,3), E(2,2). Now plug in each pairs with the equation:
F= 5x + 2y
For A(3,6), →→F=15+12 = 25
For B(0,6), →→F=0+12 = 12
For C(6,0), →→F=30+0 = 30
For D(0,3), →→F=0+6 = 6
For E(2,2), →→F=20+4 = 14
So the max value of F= 5x+2y is 30 at C(6,0)
Hope that I didn't make any mistake
Xis less than or equal to 3
30+5=35
35*8=280
280 is the original number
Answer:
x = 6
∠B = 126
Step-by-step explanation:
∠A = ∠B
where
∠A = 8x + 78°
∠B = 2x + 114°
8x + 78° = 2x + 114°
8x - 2x = 114 - 78
6x = 36
x = 36/6
x = 6
plugin x into ∠B = 2x + 114°
∠B = 2x + 114°
∠B = 2(6) + 114°
∠B = 126
