# Linear programming alternate optimum solutions

**Q- The Optimum of a linear programming problem occurs at $$(1,2,3)$$ and $$(-1,0,7)$$ then the optimum also occurs at?

$$a)(2,4,6)$$

$$b)(0,3,5)$$

$$c)(0,1,5)$$

$$d)(3,2,1)$$

$$e)$$ None of the above.

When we are given two points in the xy axis as optimum points we can simply take any point which lies on that line as other optimum solutions to the given problem.But i tried to work out the answer to this problem in the same way but i cant figure out an answer without the optimum simplex tableau.So how to find the answer to this question??**

• The direction vector can be constructed by building the difference of the two given points-multiplied with the factor r. For the Position vector you can take the Vektor (1,2,3). The sum should be equal to one of the given points: $\begin{pmatrix} 1 \\2\\3 \end{pmatrix}+r \cdot \begin{pmatrix} x_1-x_2 \\y_1-y_2\\z_1-z_2 \end{pmatrix}=\begin{pmatrix} p_1 \\p_2\\p_3 \end{pmatrix}$. The condition $|r| \leq 1$ has also be fullfilled. Commented Jun 22, 2014 at 9:54
• Your answer is correct. If the optimal solution space contains two points, then the optimal solution space is a line-in case of linear programming. So we construct a vector, which contents the two points. This can be done with the position vector and the direction vector. We don´t know how long the line is. But we know it must be at least as long that the two given points are in it. If $|r|\leq 1$, then the third point is in the optimal solution space (line). If r would be $|r| \geq 1$ the third point could be beyond the bounderies. Commented Jun 22, 2014 at 10:24