Question

optimization with linear programming
maximum and minimum values when a system of linear inequalities produces a bounded polygonal region, the maximum or minimum value of a related function will occur at a vertex of the region.
example: graph the system of inequalities. name the coordinates of the vertices of the feasible region. find the maximum and minimum values of the function ( f(x,y)=3x + 2y ) for this polygonal region.
( yleq4 )
( yleq -x + 6 )
( ygeq\frac{1}{2}x-\frac{3}{2} )
( yleq6x + 4 )
first find the vertices of the bounded region.
graph the inequalities.
the polygon formed is a quadrilateral with vertices at ( (0,4),(2,4),(5,1) ), and ( (-1,-2) ). use the table to find the maximum and minimum values of ( f(x,y)=3x + 2y ).
the maximum value is 17 at ( (5,1) ). the minimum value is -7 at ( (-1,-2) ).
exercises
graph each system of inequalities. name the coordinates of the vertices of the feasible region. find the maximum and minimum values of the given function for this region.

1. ( ygeq2 )

( 1leq xleq5 )
( yleq x + 3 )
( f(x,y)=3x - 2y )

1. ( ygeq - 2 )

( ygeq2x - 4 )
( x - 2ygeq - 1 )
( f(x,y)=4x - y )

1. ( x + ygeq2 )

( 4yleq x + 8 )
( ygeq2x - 5 )
( f(x,y)=4x + 3y )

Explanation:

Step1: Find vertices

• For \(y = 2\) and \(x = 1\), the point is \((1,2)\).
• For \(y = 2\) and \(x = 5\), the point is \((5,2)\).
• Solve \(y=x + 3\) and \(x = 1\), substitute \(x = 1\) into \(y=x + 3\), we get \(y=4\), so the point is \((1,4)\).
• Solve \(y=x + 3\) and \(x = 5\), substitute \(x = 5\) into \(y=x + 3\), we get \(y = 8\), so the point is \((5,8)\).

Step2: Calculate function values

• For \((1,2)\): \(f(1,2)=3\times1-2\times2=3 - 4=-1\).
• For \((5,2)\): \(f(5,2)=3\times5-2\times2=15 - 4 = 11\).
• For \((1,4)\): \(f(1,4)=3\times1-2\times4=3-8=-5\).
• For \((5,8)\): \(f(5,8)=3\times5-2\times8=15 - 16=-1\).

Answer:

The vertices of the feasible region are \((1,2)\), \((5,2)\), \((1,4)\), \((5,8)\). The minimum value of \(f(x,y)=3x - 2y\) is \(-5\) at \((1,4)\) and the maximum value is \(11\) at \((5,2)\).