What is the maximum value of the objective function, p, with the given constraints?
p=10x+50y
x+y≤18
x+3y≤30
x≥0
y≥0
180
420
500
900

The maximum value of the objective function is P=500

Step-by-step explanation:

we have

----> constraint A

----> constraint B

----> constraint C

----> constraint D

Solve the system of inequalities by graphing

The solution is the shaded area

see the attached figure

The vertices of the shaded area are

(0,0), (0,10),(12,6),(18,0)

Find the maximum value of the objective function

P=10x+50y

For each vertex substitute the value of x and the value of y in the objective function

1) For (0,0)

P=10(0)+50(0)=0

2) For (0,10)

P=10(0)+50(10)=500

3) For (12,6)

P=10(12)+50(6)=420

4) For (18,0)

P=10(18)+50(0)=180

therefore

The maximum value of the objective function is P=500

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !