Answer:
I can't see the picture or handwriting
Lagrange multipliers:







(if

)

(if

)

(if

)
In the first octant, we assume

, so we can ignore the caveats above. Now,

so that the only critical point in the region of interest is (1, 2, 2), for which we get a maximum value of

.
We also need to check the boundary of the region, i.e. the intersection of

with the three coordinate axes. But in each case, we would end up setting at least one of the variables to 0, which would force

, so the point we found is the only extremum.
So what we do is
area that remains=total area-triangle area that was cut out
we need to find 2 things
total area
triangle area
total area=rectange=base times height
area=(3x+4) times (2x+3)
FOIL or distribute
6x^2+8x+9x+12=6x^2+17x+12
triangle area=1/2 times base times height
triangle area=1/2 times (2x+2) times (x-2)=
(x+2) times (x-2)=x^2+2x-2x-4=x^2-4
so
total area=6x^2+17x+12
triangle area=x^2-4
subtract
area that remains=total area-triangle area that was cut out
area that remains=6x^2+17x+12-(x^2-4)=
6x^2+17x+12-x^2+4=
6x^2-x^2+17x+12+4=
5x^2+17x+16
area that remains is 5x^2+17x+16
Answer:
141 cans per day
Step-by-step explanation:
2000 goal cans - 872 collected cans = 1,128 still needed
1,128/8 days = 141 cans per day
Using a scientific calculator:
log 12 = 1.079181246
Rounded to four decimal places:
log 12 = 1.0792
Characteristic: 1; Mantis: 0.0792
Answer: OPtion B. 1.0792; 1; 0.0792