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Question 821752: solve where R=10x+15y subject to 2x+3y<= 24, 3x+2y<=36, x>=0, y>=0
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website! - Each of the four inequalities has an associated equation:
2x + 3y = 24, 3x + 2y = 36, x = 0 and y = 0 - The graphs of these equations are lines.
- x = 0 is the equation for the y-axis and y = 0 is the equation for the x-axis.
- Graph the other two lines on the same graph.
- The two lines graphed above plus the x and y axes should form some kind of quadrilateral in the first quadrant.
- The points inside and along the edges of this quadrilateral are all the possible solutions which fit all four inequalities.
- The quadrilateral has four vertices:
- The origin, (0, 0)
- The point where the line 3x + 2y = 36 intersects the x-axis.
- The point where the line 2x + 3y = 24 intersects the y-axis.
- The point where the lines 2x + 3y = 24 and 3x + 2y = 36 intersect each other.
- Find the three unknown vertices.
- Among all the solutions (the points along the edges and the points in the interior of the quadrilateral), the largest (and smallest) possible values for R will come from the coordinates of one of the vertices of the quadrilateral. So try the coordinates of each vertex, one vertex at a time, in the equation for R: R = 10x + 15y. This will result in an R for each vertex. One of them will be the largest possible value for R and one of them will be the smallest possible value for R. (Your post did not say which you were looking for, the largest/maximum or smallest/minimum value for R.)
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