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  • Minimise Z equals 13x – 15y subject to the constraints: x plus y less than equals 7, 2x – 3y plus 6 greater than equals 0, x greater than equals 0, y greater than equals 0.

Minimise Z = 13x -15y  subject to the constraints: x + y \leq 7, 2x - 3y + 6 \geq 0, x \geq 0, y \geq 0.\\

Answers (1)

It is given that:

Z = 13x -15y

It is subject to constraints

x + y $ \leq $ 7, 2x - 3y + 6 $ \geq $ 0, x $ \geq $ 0, y $ \geq $ 0.\\

Now let us convert the given inequalities into equation
We obtain the following equation

\\ x+y \leq 7 \\ \Rightarrow x+y=7 \\ 2 x-3 y+6 \geq 0 \\ \Rightarrow 2 x-3 y+6=0 \\ y \geq 0 \\ \Rightarrow x=0 \\ y \geq 0 \\ y=0

The region which is represented by x+y$ \leq $ 7:\\

The line in the sum meets the coordinate axes (7,0) and (0,7) respectively. If we join the lines, we will get the other line that is x + y =7. And then it is further clear that (0,0) satisfies the inequation. Then the origin represents the solution for the set of the inequation x+y$ \leq $ 7:\\

The region represented by 2x - 3y + 6 $ \geq $ 0:\\

The line 2x-3y+6=0 collides with the other axes to coordinate (-3,0) and (0,2) respectively. Then the lines are joined further to obtain the line 2x-3y+6=0. So, the part that contains the origin then represents the other solution set of the inequation 2x - 3y + 6 $ \geq $ 0. \\ \\

Looking at the graph we get, 

The shaded region OBCD is the feasible region is bounded, so, minimum value will occur at a corner point of the feasible region.
Corner Points are O(0,0), B(0,2), C(3,4) and D(7,0)
Now we will substitute these values in Z at each of these corner points, we get

\begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=13 x-15 y \\ \hline O(0,0) & Z=13(0)-15(0)=0+0=0 \\ B(0,2) & Z=13(0)-15(2)=0-30=-30 \rightarrow \min \\ C(3,4) & Z=13(3)-15(4)=39-60=-21 \\ D(7,0) & Z=13(7)-15(0)=91-0=91 \end{array}

So, the final answer of the question is the minimum value of Z is -30 at the point of (0,2).

 

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