Find the maximum value of the objective function f(x, y) = x − y + 1, subject to the constraints x ≥ 0, − x + y ≥ 0, and y ≤ 2.
we see that
x≥0
-x+y≥0 or y≥x
and
y≤2
we see
x-y+1
y is being minused, so make it very small, we can make it 0
because y can equal x
and x can equal 0
so
f(0,0)=0-0+1=1
at (0,0) yeilds a value of 1
sorr if wrong
The maximum value of the objective function is 1 if the subject to the constraints x ≥ 0, − x + y ≥ 0, and y ≤ 2.
It is defined as the expression in mathematics in which both sides are not equal they have mathematical signs either less than or greater than known as inequality.
We have:
The objective function f(x, y) = x − y + 1
The subject to constraints:
x ≥ 0, − x + y ≥ 0, and y ≤ 2
First, plot all the inequality, the coordinate plane:
The intersection region is shown in the graph.
The objective function:
f(x, y) = x − y + 1
Checking boundary points:
Plug x = 0 and y = 2
f(0, 2) = 0 − 2 + 1
f(0, 2) = -1
Plug x = 2 and y = 2
f(2, 2) = 2 − 2 + 1
f(2, 2) = 1
Plug x = 0 and y = 0
f(0, 0) = 0 − 0 + 1
f(0, 0) = 1
Plug x = 1 and y = 1
f(1, 1) = 1 − 1 + 1
f(1, 1) = 1
Plug x = 0 and y = 1
f(0, 1) = 0 − 1 + 1
f(0, 1) = 0
Plug x = 1 and y = 2
f(1, 2) = 1 − 2 + 1
f(1, 2) = 0
Thus, the maximum value of the objective function is 1 if the subject to the constraints x ≥ 0, − x + y ≥ 0, and y ≤ 2.
Learn more about the inequality here:
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