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 and Step-by-step explanation:

This is the answer.
<span>8 - 3x - 5x = -8
8 - 8x = -8
-8x = -8 - 8
-8x = -16
8x = 16
x = 16/8
x = 2</span>
<h3>
Answer: 3x^3 + 11x^2 - 5x - 25</h3>
Work Shown:
yz = y(x^2+2x-5)
yz = y(x^2) + y(2x) + y(-5)
yz = x^2( y ) + 2x( y ) - 5( y )
yz = x^2( 3x+5 ) + 2x( 3x+5 ) - 5( 3x+5 )
yz = x^2*3x + x^2*5 + 2x*3x + 2x*5 - 5*3x - 5*5
yz = 3x^3 + 5x^2 + 6x^2 + 10x - 15x - 25
yz = 3x^3 + 11x^2 - 5x - 25