Question

Graph the system of constraints and find the value of x and y that maximize the objective function.. x>0. y>0. y<1/5x+2. 5>y+x. objective function: c=7x-3y

Answer 1

X > 0 , everything to the right of the y-axis 
y > 0 , everything above the x-axis 
So far you are in quadrant I of the x-y plane
y<1/5x+2
sketch a line leaning 45° to the left with a y-intercept of 2, shade in everything below that line 
now our Objective Function is 
C = 7x - 3y , which is a line with slope 7/3 
letting that line go through the origin, let it slide parallel to itself over the shaded region until you reach the farthest point from the origin. 
Here are the required values : 
c = 35; x = 5, y = 0

Answer 2

Answer: The value of x= 5

And the value of y =0.

Step-by-step explanation:

Since we have given that

$Max\ c=7x-3y\\\\yy+x\\\\s.t.\ x>0\ y>0$

As we can see from the graph, there are three points occurred in feasible region, which are as follows;

$A(2.5,2.5)\\\\B(-10,0)\\\\C(5,0)$

Now, we put the above three values in the objective function:

$c=7x-3y\\\\Put\ A(2.5,2.5)\\\\c=7\times 2.5-3\times 2.5\\\\c=17.5-7.5\\\\c=10\\\\Similarly,\\\\Put\ B(-10,0)\\\\c=7\times -10-3\times 0\\\\c=-70\\\\Similarly,\\\\Put\ C(5,0)\\\\c=7\times 5-3\times 0\\\\c=35$

Therefore, we can see that

At C(5,0) we get maximum value as c = 35.

Hence,  The value of x= 5

And the value of y =0.