Answer:
The point that maximizes the objective function is (3,0)
Step-by-step explanation:
we have
Constraints:

Using a graphing tool
The feasible region is the shaded area
see the attached figure
The vertices of the feasible region are
(0,0),(0,1),(1.5,1.5) and (3,0)
we know that
To find the point in the feasible region that maximizes the objective function, replace each ordered pair of vertices in the objective function and then compare the results.
The objective function is

For (0,0) -----> 
For (0,1) -----> 
For (1.5,1.5) -----> 
For (3,0) -----> 
therefore
The point that maximizes the objective function is (3,0)
If the numerator and the denominator do not have a greatest common factor then the faction is in simplest form
Answer:
option 1.
Step-by-step explanation:
Answer:

Step-by-step explanation:
We want to prove algebraically that:

is a parabola.
We use the relations

and

Before we substitute, let us rewrite the equation to get:

Or

Expand :

We now substitute to get:

This means that:

Square:

Expand:




This is a parabola (0,2.5) and turns upside down.