Question

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.   
14. 5 ≥ y ≥ ­3 
4x + y ≤ 5  ­2x + y ≤ 5 
f(x, y) = 4x ­ 3y 

15. x ≥ ­10 
1 ≥ y ≥ ­6 
3x + 4y ≤ ­8 
2y ≥ x ­ 10 
f(x, y) = 2x + y 

Answer 1

ANSWER TO QUESTIONS 14

We need to graph all the given constraints(inequalities). The intersection of the of all the graphed system of inequalities gives the vertices of the feasible region.

The first inequality is $5\geq y\geq -3$.  

This is very simple to graph. It is the region bounded by the horizontal lines,

$y=5$ and $y=-3$.

The next constraint or inequality is $4x+y\leq 5$

To graph this inequality, we need to graph the corresponding linear equation $4x+y=5$.  

We plot the intercepts $(0,5)$ and $(\frac{5}{4},0)$.

We then draw a solid line through the points.

Next, we test the origin to determine which half plane to shade.

$4(0)+0\leq 5$

$0\leq 5$

This statement is true so we shade the lower half plane.

We now graph the inequality,

$2x+y\leq 5$

To graph this inequality, we need to graph the corresponding linear equation $-2x+y=5$.  

We plot the intercepts $(0,5)$ and $(-\frac{5}{2},0)$.

We then draw a solid line through the points.

Next, we test the origin to determine which half plane to shade.

$-2(0)+0\leq 5$

$0\leq 5$

This statement is true so we shade the lower half plane.

The feasible region is shown in the above diagram.

The vertices of the feasible region are;

$A(0,5),B(2,-3),D(-4,-3)$

We substitute the vertices into the objective function to obtain;

$f(x,y)=4x-3y$

$A(0,5)$ gives ;$f(0,5)=4(0)-3(5)=-15$

$B(2,-3)$ gives ;$f(2,-3)=4(2)-3(-3)=17$

$D(-4,-3)$ gives ;$f(-4,-3)=4(-4)-3(-3)=-7$

Therefore the maximum value is $17$ and it occurs at $(2,-3)$.

The minimum value is $-15$ and it occurs at $(0,5)$.

ANSWER TO QUESTION 15

We repeat the same process.

To graph $x\geq -10$, we graph the corresponding linear equation,

$x=-10$ and shade the right half plane.

The graph of $1\geq y\geq -6$ is the region bounded between $y=1$ and $y=-6$.

To graph $3x+4y\leq 8$, we graph the corresponding linear equation, $3x+4y=8$.

After testing the origin we shade the lower half plane.

Last but not the least, we graph $2y \geq x-10$ but graphing the corresponding linear equation with solid line and shading the upper half plane after testing for the origin.

The vertices of the feasible region are $A(-4,1),B(2.4,-3.8),C(-2,-6),D(-10,-6),(-10,1)$.  

We substitute in to the objective function to get  

$P(x,y)=2x+y$,

At $A(-4,1)$, $P(-4,1)=2(-4)+1=-7$,

At $B(2.4,-3.8)$, $P(2.4,-3.8)=2(2.4)-3.8=1$.

At $C(-2,-6)$, $P(-2,-6)=2(-2)-6=-10$.

At $D(-10,-6)$, $P(-10,-6)=2(-10)-6=-26$.

At $E(-10,1)$, $P(-10,1)=2(-10)+1=-19$,

The maximum value is 1 and it occurs at $(2.4,-3.8)$.

The minimum value is $-26$ and it occurs at $(-10,-6)$