Question

the graph to the right shows a region of feasible solutions. use this region to find maximum and minimum values of the given objective functions, and the locations of these values on the graph. (a) $z = 0.40x + 1.25y$ (b) $z = 2.50x + 0.25y$ boxes to complete your choice. \\(\bigcirc\\) a. the maximum value of the objective function $z = 2.50x + 0.25y$ is \\(\square\\) and is located at \\(\square\\). (type an exact answer.) \\(\bigcirc\\) b. the maximum does not exist. what is the minimum of $z = 2.50x + 0.25y$? select the correct answer below and, if necessary, fill in the answer boxes to complete your choice.

Explanation:

Step1: Identify Vertices

First, we need to find the vertices of the feasible region. From the graph, we can see the vertices are \((0,0)\), \((0,11)\), and we need to find the third vertex. Wait, the graph seems to be a polygon, but maybe it's a triangle or a quadrilateral? Wait, the given points are \((0,0)\) and \((0,11)\), and let's assume the other vertex is \((x,0)\) but no, wait, the feasible region is a polygon. Wait, maybe the third vertex is where the line from \((0,11)\) meets the x - axis? Wait, no, the graph is a blue region. Wait, maybe the feasible region has vertices at \((0,0)\), \((0,11)\), and let's say \((a,0)\), but actually, in linear programming, the maximum and minimum of a linear objective function over a convex polygon (feasible region) occur at the vertices.

Wait, the objective function is \(z = 2.50x+0.25y\). Let's list the vertices. From the graph, we can see two vertices: \((0,0)\) and \((0,11)\), and we need to find the third vertex. Wait, maybe the right - most vertex? Wait, the graph has a point \((0,11)\) and \((0,0)\), and let's assume the other vertex is \((x,0)\), but actually, maybe the feasible region is a triangle with vertices \((0,0)\), \((0,11)\), and \((c,0)\), but we need to check. Wait, maybe the graph is a polygon where the third vertex is, for example, if we assume the line from \((0,11)\) has a slope, but maybe the problem is that the feasible region is unbounded? Wait, no, the blue region looks bounded. Wait, maybe the vertices are \((0,0)\), \((0,11)\), and let's say \((n,0)\), but actually, in the graph, maybe the third vertex is \((x,0)\), but we need to calculate \(z\) at each vertex.

Step2: Evaluate \(z\) at Vertices

Let's assume the vertices are \((0,0)\), \((0,11)\), and \((k,0)\). Wait, but maybe the feasible region is a triangle with vertices \((0,0)\), \((0,11)\), and let's say \((a,0)\). But wait, maybe the graph is such that the feasible region is a polygon with vertices \((0,0)\), \((0,11)\), and \((x,0)\). Wait, no, maybe the other vertex is where the line from \((0,11)\) meets the x - axis, but we need to check. Wait, maybe the feasible region is unbounded? Wait, the objective function \(z = 2.50x + 0.25y\) has a positive coefficient for \(x\) and positive for \(y\). If the feasible region is unbounded in the direction of increasing \(x\), then the maximum would not exist. Wait, let's check the coefficients. The coefficient of \(x\) is \(2.50\) which is positive. If the feasible region is unbounded in the positive \(x\) - direction (i.e., we can take \(x\) as large as we want while still being in the feasible region), then as \(x\) increases, \(z=2.50x + 0.25y\) will also increase without bound (since \(2.50>0\)). So, in that case, the maximum does not exist.

For the minimum: We evaluate \(z\) at the vertex \((0,0)\): \(z(0,0)=2.50\times0 + 0.25\times0=0\). At \((0,11)\): \(z(0,11)=2.50\times0+0.25\times11 = 2.75\). If the feasible region is unbounded in the positive \(x\) - direction, but for the minimum, since the coefficient of \(x\) and \(y\) are non - negative, the minimum occurs at \((0,0)\) because as \(x\) or \(y\) increase, \(z\) increases (since \(2.50>0\) and \(0.25>0\)).

Step3: Analyze Maximum

Since the coefficient of \(x\) in \(z = 2.50x+0.25y\) is positive (\(2.50>0\)), and if the feasible region is unbounded in the positive \(x\) - direction (i.e., we can move along the positive \(x\) - axis infinitely while staying in the feasible region), then as \(x\) approaches \(+\infty\), \(z = 2.50x+0.25y\) will also approach \(+\infty\) (because \(2.50x\) dominates as \(x\) becomes large). So, the maximum value of \(z\) does not exist.

Step4: Analyze Minimum

To find the minimum, we evaluate \(z\) at the vertices of the feasible region. The vertices are \((0,0)\), \((0,11)\), and if there are other vertices, but since the objective function \(z = 2.50x + 0.25y\) has non - negative coefficients, the smallest value of \(z\) will occur at the vertex with the smallest \(x\) and \(y\) values. At \((0,0)\): \(z=2.50\times0 + 0.25\times0 = 0\). At \((0,11)\): \(z=2.50\times0+0.25\times11=2.75\). So, the minimum value of \(z\) is \(0\) and it occurs at \((0,0)\).

Answer:

For the maximum of \(z = 2.50x+0.25y\): B. The maximum does not exist.

For the minimum of \(z = 2.50x+0.25y\): The minimum value is \(0\) and it is located at \((0,0)\).