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The maximum value of the objective function is 26 and the minimum is -10
<h3>How to determine the maximum and the minimum values?</h3>The objective function is given as:
z=−3x+5y
The constraints are
x+y≥−2
3x−y≤2
x−y≥−4
Start by plotting the constraints on a graph (see attachment)
From the attached graph, the vertices of the feasible region are
(3, 7), (0, -2), (-3, 1)
Substitute these values in the objective function
So, we have
z= −3 * 3 + 5 * 7 = 26
z= −3 * 0 + 5 * -2 = -10
z= −3 * -3 + 5 * 1 =14
Using the above values, we have:
The maximum value of the objective function is 26 and the minimum is -10
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$=(a^2-10a)-(b^2+6b) +16$
$=[(a^2-2(5)a+25)-25]-[(b^2+2(3)b+9)-9]+16$
$=(a-5)^2-25-(b+3)^2+9+16$
$=(a-5)^2-(b+3)^2$