# Minimum and maximum values of the objective quantity?

Hi,

I am not sure how to solve the following high school algebra problem.
As shown in the attached file, I have plotted the first three equations, but I don't know what to do with the fourth equation.

Find the minimum and maximum values of the objective quantity.
2X + Y ≤ -8
3X - 2Y ≥ -5
X - 3Y ≤ 10
f(X, Y)= 10X - 8Y

1. What are we trying to achieve in this problem?
2. How do I plot the fourth equation? I am confused by the fact that the equation is different than the other three equations.

Thanks.

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arildno
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2. You are to find out WHERE it has its maximum value, on the domain specified by your inequalities.
3. Since f is a linear function, its maxima/minima must lie on the boundary of the region, and, furthmore, at the CORNERS of your region.

How might you find those?

PS:
AS for plotting f(x,y) as a GRAPH, you would need the third dimension z=f(X,Y) to do that properly. You can consider f(X,Y) to be the local height of the slanted roof above the domain in the (x,y)-plane specified by the three inequalities.

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Ray Vickson
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Hi,

I am not sure how to solve the following high school algebra problem.
As shown in the attached file, I have plotted the first three equations, but I don't know what to do with the fourth equation.

Find the minimum and maximum values of the objective quantity.
2X + Y ≤ -8
3X - 2Y ≥ -5
X - 3Y ≤ 10
f(X, Y)= 10X - 8Y

1. What are we trying to achieve in this problem?
2. How do I plot the fourth equation? I am confused by the fact that the equation is different than the other three equations.

Thanks.
Look at http://faculty.northgeorgia.edu/kmelton/LPgraph.htm [Broken] for example.

Basically, you want to find the largest or smallest value of f(x,y) that is obtainable within the feasible region, so look at the iso-objective lines f(x,y) = constant, for various values of the constant. For example, look at the two lines f(x,y) = 1 and f(x,y) = 2. Do they intersect the feasible region? If so, what about f(x,y) = 3? What about f(x,y) = 4? In this way you will have a series of parallel lines, and you need to find that line f(x,y) = v having the largest value of the constant v (for the max problem) or the smallest value of v (for the min problem). Draw two or three of the lines, and you will soon see what is going on.

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Thank you. Got it.

arildno
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Ray Vickson gives you a very good strategy to understand why it is the CORNERS of the boundary that will provide the extrema.

Here is another one, based on the fact that f(X,Y) is linear, and when you therefore look at the attained heights of f(X,Y) relative to a STRAIGHT LINE L in the x(,y)-plane, you see that f(X,Y) is either a) strictly increasing, or b) decreasing or c) strictly constant along that line. Thus, since the boundaries are ALSO straight lines, the value of f(X,Y) on the straight line L attained will be both an extremum relative to L, since at the boundary, we have travelled longest along L (Meaning: We may therefore ignore interior points for global extrema!) AND the boundary lines are themselves valid choices for L. Since the corners are the extreme values we can go to along any of the boundary lines, that is where the global extrema may be found.(For Vickson's lines of strictly constant values of f(x,y), the value of f AT the boundary line will generally not be an extremum, unless they hit the corners)
--
Note that MY approach is, in a manner of speaking "90 degrees shifted" relative to Ray Vickson's approach:
While he looks at how f(X,Y)="constant" behaves, I look (primarily) at how f(x,y)="strictly changing" behaves.
We get the same result either way. Last edited: