The maximum value of the objective function is 330
<h3>How to maximize the
objective function?</h3>
The given parameters are:
Max w = 5y₁ + 3y₂
Subject to
y₁ + y₂ ≤ 50
2y₁ + 3y₂ ≤ 60
y₁ , y₂ ≥ 0
Start by plotting the graph of the constraints (see attachment)
From the attached graph, we have:
(y₁ , y₂) = (90, -40)
Substitute (y₁ , y₂) = (90, -40) in w = 5y₁ + 3y₂
w = 5 * 90 - 3 * 40
Evaluate
w = 330
Hence, the maximum value of the function is 330
Read more about objective functions at:
brainly.com/question/26036780
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<span>6d = 108
You would first divide each side by 6 to isolate the variable.
6d = 108
6 6
d = 18
-Gives you thumbs up- </span>
Answer:
14 is D but I don't 15 question answer
There are six p's, and those six p's have a total sum of thirty six
6p=36
Divide both sides by six to get the value of p
36/6 is 6
p is equal to six