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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Answer:
cant read
Step-by-step explanation:
:P
I think it’s yellow. Sorry if I’m wrong.
Answer:
Step-by-step explanation:
Volume of a cube:
V = a3
Surface area of a cube:
the area of each face (a x a) times 6 faces
S = 6a2
Face diagonal of a cube:
By the pythagorean theorem we know that
f2 = a2 + a2
Then f2 = 2a2
solving for f we get
f = a√2
Diagonal of the solid cube:
Again, by the pythagorean theorem we know that
d2 = a2 + f2
substituting f into this equation we get
d2 = a2 + (a√2)2 = a2 + 2a2 = 3a2
solving for d we get
d = a√3
