Mathematics - Online Test
The normal to the curve 2 y = 3 – at (1, 1) is
Objective function is Z = x + 2 y ……………………(1).
The given constraints are : x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200; x, y ≥ 0.
The corner points are obtained by drawing the lines x+2y=100, 2x-y=0 and 2x+y=200.
The points so obtained are (0,50),(20,40), (50,100) and (0,200)
Corner points | Z = x + 2y |
D(0 ,50 ) | 100……………..(Min.) |
A(20,40) | 100……………………..(Min.) |
B(50,100) | 250 |
C(0,200) | 400 |
Here , Z = 100 is minimum at ( 0, 50) and ( 20 ,40).
Minimum Z = 100 at all the points on the line segment joining the points (0, 50) and (20, 40).
then D is equal to
Equation of the line which is perpendicular to the given line x + y = 6 is x - y + k = 0
Since it passes through the point (4,3)
4 - 3 + k = 0
Therefore k = -1
Hence the equation of the line segment PQ is x - y - 1 = 0
On solving these two lines we get x = 7/2 and y = 5/2
This point of intersection is the midpoint of the line segement PQ
That is = 7/2 Hence x = 3
Similarly = 5/2. Hence y = 2
Hence the coordiates of the point Q is (3,2)
