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Given the system of inequalities below, determine the shape of the feasible region anc d the corner points of the feasible region. Give the shape as "triangle", "quadriateral", or "unbounded". Report your cornor points starting with the one which has the smallest x-value. If more than one comer point has the same smallest x-value, start with the one that has the smallest y-value. Proceed clockwise from the first corner point. Leave any unnecessary answer spaces blankx+y≥7 6x+y≤10

User Slothrop
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Final answer:

The feasible region formed by the system of inequalities x + y ≥ 7 and 6x + y ≤ 10 is a triangle, with the corner points being (0,7), (1,6), and (1.67,0) in a clockwise direction.

Step-by-step explanation:

To determine the shape of the feasible region and the corner points of the system of inequalities x + y ≥ 7 and 6x + y ≤ 10, we will first graph these inequalities on the Cartesian plane.

The inequality x + y ≥ 7 is the region above or on the line when x + y = 7. The inequality 6x + y ≤ 10 is the region below or on the line when 6x + y = 10. To find the boundary lines, we can set up a table of values or find the intercepts for each equation.

For x + y = 7 (boundary of the first inequality), the intercepts would be (0,7) on the y-axis and (7,0) on the x-axis. Similarly, for 6x + y = 10 (boundary of the second inequality), the intercepts would be (0,10) on the y-axis and (10/6, 0) or approximately (1.67, 0) on the x-axis.

By graphing these lines and shading the appropriate regions, we observe that the feasible region is a triangle. The corner points of this triangle where the lines intersect are found by solving the system of equations x + y = 7 and 6x + y = 10 simultaneously.

Solving this system gives us the coordinate (1,6) as the intersection point. Therefore, the corner points, starting with the smallest x-value, are (0,7), (1,6), and (1.67,0), proceeding in a clockwise direction.

User Ozkolonur
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