Max w = 5y1 + 3y2 s. a. y1 + y2 ≤ 50 2y1 + 3y2 ≤ 60 y1 , y2 ≥ 0
1 answer:
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
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Let the cost of gasoline in the year 2000 be represented b the equation
y = a + b*x
where
x = months, counted from January
y = cost, dollars
The given data in the table is
Month: Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
x, months: 1 2 3 4 5 6 7 8 9 10 11 12
y, dollars: --- --- --- --- 1.76 2.13 --- --- --- --- --- ---
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a + 5b = 1.76 (1)
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a + 6b = 2.13 (2)
Subtract equation (1) from (2).
a + 6b - (a + 5b) = 2.13 - 1.76
b = 0.37
From (1), obtain
a = 1.76 - 5b
= 1.76 - 5*0.37
= -0.09
The required equation is
y = 0.37x - 0.09
The graph shows the line, with the given data for May and June.
Answer: D. y = 0.37x - 0.09
y = a + b*x
where
x = months, counted from January
y = cost, dollars
The given data in the table is
Month: Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
x, months: 1 2 3 4 5 6 7 8 9 10 11 12
y, dollars: --- --- --- --- 1.76 2.13 --- --- --- --- --- ---
When x = 5, y = 1.76.
Therefore
a + 5b = 1.76 (1)
When x = 6, y = 2.13
Therefore
a + 6b = 2.13 (2)
Subtract equation (1) from (2).
a + 6b - (a + 5b) = 2.13 - 1.76
b = 0.37
From (1), obtain
a = 1.76 - 5b
= 1.76 - 5*0.37
= -0.09
The required equation is
y = 0.37x - 0.09
The graph shows the line, with the given data for May and June.
Answer: D. y = 0.37x - 0.09