For the differential equationfind the solution curve passing through the point (1, –1).
Let number of food type P = x
And number of units of food type Q = y
Therefore , the above L.P.P. is given as :
Minimise , Z = 60x +80y , subject to the constraints : 3 x + 4y ≥ 8, 5x + 2y ≥ 11, x,y ≥ 0.
The corner points can be obtained by drawing the lines 3x+4y=8 and 5x+2y=11 graphically.
The points so obtained are (8/3,0), (2,1/2), (0,11/2)
Corner points | Z = - x + 2y |
D(8/3,0 ) | 160………………….(Min.) |
A(2,1/2) | 160………………(Min.) |
B(0,11/2) | 440 |
Here Z = 160 is minimum. i.e. Minimum cost = Rs 160 at all points lying on segment joining .
The area bounded by the curves is given by