Which term best describes the solution of the situation represented by the system of inequalities? (Assume that x is greather th
an or equal to 0 and y is greater than or equal to 0)
3x+4y is less than or eaual to 8
2x+4y is less than or equal to 6
f(x,y)=x+y
A. alternate optimal solutions
B. One Optimal Solution
C. Unbounded
D. Infeasible
3x+4y is less than or eaual to 8
2x+4y is less than or equal to 6
f(x,y)=x+y
A. alternate optimal solutions
B. One Optimal Solution
C. Unbounded
D. Infeasible
2 answers:
The system of Inequality given here is
1. →x≥0
2.→y≥0
3.→3x+4y≤8
4.→2x+4y≤6
f(x,y)=x+y
Drawing the Inequality on the Coordinate plane and obtaining the feasible points
And Feasible Points are
1.→(0,1.5)
2.→(0,0)
3.→(2.667,0)
4.→(2,0.5)
→f(0,1.5)=0+1.5=1.5
→f(0,0)=0+0=0
→f(2.667,0)=2.667+0=2.667
→f(2,0.5)=2+0.5=2.5
⇒There is one Optimal Solution obtained at Point (2.667,0).
Option B
One optimal solutions
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Answer: 
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So we know that there is a restriction on the y-axis.
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As x approaches the asymptote of 7⁻, then y would be diverging out to negative infinite.
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So, our range would be:
Explanation: For this, it is often best to find the horizontal asymptote, and then take limits as x approaches the vertical asymptote and the end behaviours.
Well, we know there will be a horizontal asymptote at y = 0, because as x approaches infinite and negative infinite, the graph will shrink down closer and closer to 0, but never touch it. We call this a horizontal asymptote.
So we know that there is a restriction on the y-axis.
Now, since we know the end behaviours, let's find the asymptotic behaviours.
As x approaches the asymptote of 7⁻, then y would be diverging out to negative infinite.
As x approaches the asymptote at 7⁺, then y would be diverging out to negative infinite.
So, our range would be: