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

where's the diagrams?

the answer is:

the answer is:

to solve the problem, we need to remember the quotient of power property, it's defined by the following relation:

if we have a quotienf of powers that have the same base, we need to keep the same base and subtract the exponent of the denominator power to the exponent of the numerator power.

so, we are given the expression:

then, calculating we have:

hence, the answer is:

have a nice day!