College Algebra

Question

Mmaximize the function f(x y)=2x+3y in the region determined by the following constraints. 3x+2y<18. 3x+4y>12. X>0. Y>0

Answer 1

the intersection of 3x+2y=18 vs 3x+4y=12 is -2y = 6 upon subtracting.

So y=-3 --> x=8; the intersection is (8,-3) which is not in the 1st quadrant

The region has vertices that are the intercepts: (0,3) (0,9), (6,0),(4,0)

Plugging these into the objective function:

f(0,3) = 9, f(0,9) = 27, f(6,0)=12, and f(4,0) = 8

So your winner is (0,9)

Answer 2

Start by graphing the two constraints.

(Which I can not do here.)

GRAPHING TIPS:

> For each constraint, graph two points.

Graph the x and y intercepts.

For the first constraint, you get: (0,9), (6,0)

For the 2nd constraint, you get: (0,3), (4,0)

> Drawing a dotted line between the two sets of intercepts

and shade in the feasible region.

> Note that x > 0 and y > 0

Those are constraints too. But these constraints

just tell you to stay in the first quadrant

where both x and y are positive.

> With all four constraints, you will have a

4-sided region. The four corners are the

critical points.

FIND THE EXTREMA

> To find the extrema, evaluate the function

at each critical point.

f(x,y) = 2x + 3y

f(0.9) = 2(0) + 3(9) = 27. MAX

f(6,0) = 2(6) + 3(0) = 12

f(0,3) = 2(0) + 3(3) = 9

f(4,0) = 2(4) + 3(0) = 8

Note: The feasible region

does NOT include the lines and corners

because the equations for the constraints are

inequalities without the equal part.

I have a feeling that the original problem statement

showed constraints with inequalities WITH the equal part.

If the problem statement is correct and the constraints

are correct, then the MAX would be < 27.