Consider the feasible region in the xy-plane defined by the following linear inequalities. x≥0
y ≥0
x ≤ 10
x +y≥ 5
x + 2y ≤ 18
Part 2 Exercises:
1. Find the coordinates of the vertices of the feasible region. Clearly show how each vertex is determined and which lines form the vertex.
2. What is the maximum and the minimum value of the function Q = 60x+78y on the feasible region?

1. (0,5), (0,9), (10,4), (10,0), (5,0)

2.

Step-by-step explanation:

1.

In order to determine the coordinates of the vertices of the feasible region, we must first graph each of the inequalities. The feasible region is the region where all the inequalities cross each other. In this case it's the region shaded on the attached picture.

The first point is the intercept between the equations x=0 and x+y=5 so in order to find this first coordinate we need to substitute x=0 and solve for y.

0+y=5

y=5

(0,5)

The next point is the intercept between the equations x=0 and x+2y=18, so again, we substitute x for zero and solve for y:

0+2y=18

y=9

(0,9)

The next coordinate is the intercept between the lines x=10 and x+2y=18, so we substitute x for 10 and solve for y:

10+2y=18

2y=18-10

2y=8

y=4, so the oint is

(10,4)

The next point is the intercept between the lines x=10 and y=0, so the point is:

(10,0)

The final point is the intercept between the equations: y=0 and x+y=5. We substitute y for zero and solve for x:

x+0=5

x=5

so the point is:

(5,0).

2. In order to determine the maximum and minimum value of the function Q=60x+78y on the feasible region, we must evaluate it for each of the points found on part 1.

(0,5)

Q=60(0)+78(5)

Q=390

(0,9)

Q=60(0)+78(9)

Q=702

(10,4)

Q=60(10)+78(4)

Q=912

(10,0)

Q=60(10)+78(0)

Q=600

(5,0)

Q=60(5)+78(0)

Q=300

So now we compare the answers and pick the minimum and maximum results.

We get that:

when x=10 and y=4

and

When x=5 and y=0

step-by-step explanation:

x = central angle apc = measure of minor arc ac

the two angles x and 30 add to 180, so x+30 = 180 and x = 180-30 = 150. the measure of minor arc ac is 150 degrees.

what we need to do from here is find the circumference and then multiply it by x/360 since there are 360 degrees in a circle. in this case, x = 150, so x/360 = 150/360 = 15/36 = 5/12. the shortest distance along the curve from a to c is 5/12 of the distance around the entire circle (aka circumference)

c = circumference

c = pi*d

c = pi*24

c = 24pi

now multiply this by 5/12 to get

distance along minor arc ac = (5/12)*(circumference)

distance along minor arc ac = (5/12)*(24pi)

distance along minor arc ac = 10pi

the formula you can use is

l = (x/360)*pi*d

with

l = length of minor arc

x = central angle

d = diameter

which essentially is the steps we took though a bit differently.