Question

Please help me with this. four potential solutions.450, 780, 647, 354

Answer 1

So first of all let's take:

`x_1=x\text{ and }x_2=y`

Then we get:

`\begin{gathered} \text{Min}z=1.5x+2y \\ x+y\ge300 \\ 2x+y\ge400 \\ 2x+5y\leq750 \\ x,y\ge0 \end{gathered}`

The next step would be operate with the inequalities and the equation so we end up having only the term y at the left side of each:

`\begin{gathered} \text{Min}z=1.5x+2y \\ 1.5x+2y=\text{Min}z \\ 2y=\text{Min}z-1.5x \\ y=\frac{\text{Min}z}{2}-0.75x \end{gathered}`
`\begin{gathered} x+y\ge300 \\ y\ge300-x \end{gathered}`
`\begin{gathered} 2x+y\ge400 \\ y\ge400-2x \end{gathered}`
`\begin{gathered} 2x+5y\leq750 \\ y\leq150-(2)/(5)x \end{gathered}`

So now we have the following inequalities and equality:

`\begin{gathered} y=\frac{\text{Min}z}{2}-0.75x \\ y\ge300-x \\ y\ge400-2x \\ y\leq150-(2)/(5)x \end{gathered}`

If we take the three inequalities and replace their symbols by "=' we'll have three equations of a line:

`\begin{gathered} y=300-x \\ y=400-2x \\ y=150-(2)/(5)x \end{gathered}`

The following step is graphing these three lines and delimitating a zone in the grid that meets the inequalities:

Where the blue area is under the graph of y=150-(2/5)x which means that it meets:

`y\leq150-(2)/(5)x`

And it is also above the x-axis, y=400-2x and y=300-x which means that it also meets:

`\begin{gathered} x\ge0 \\ y\ge0 \\ y\ge400-2x \\ y\ge300-x \end{gathered}`

All of this means that the values of x and y that give us the correct minimum of z are given by the coordinates of a point inside the blue area. The next thing to do is take the four possible values for Min(z) and use them to graph four lines using this equation:

`y=\frac{\text{Min}z}{2}-0.75x`

Then we have four equations of a line:

`\begin{gathered} y=(450)/(2)-0.75x \\ y=(780)/(2)-0.75x \\ y=(647)/(2)-0.75x \\ y=(354)/(2)-0.75x \end{gathered}`

The line that has more points inside the blue area is the one made with the closest value to Min(z). Then we have the following graph:

As you can see there are two lines that have points inside the blue area. These are:

`\begin{gathered} y=-(3)/(4)x+(450)/(2) \\ y=-(3)/(4)x+(354)/(2) \end{gathered}`

That where made using:

`\begin{gathered} \text{Min }z=450 \\ \text{Min }z=354 \end{gathered}`

Taking a closer look you can see that the part of the orange line inside the blue area is larger than that of the red line. Then the value used to make the orange line would be a better aproximation for the Min z. The orange line is -(3/4)x+450/2 which means that the answer to this problem is the first option, 450.