Mathematics - Online Test

Let number of bags of cattle feed of brand P = x
And number of bags of cattle feed of brand Q = y
Therefore , the above L.P.P. is given as :
Minimise , Z = 250x +200y , subject to the constraints : 3 x + 1.5y ≥ 80, 2.5x + 11.25y ≥ 45, 2x + 3y ≥ 24 , x,y ≥ 0.,

Here Z = 1950 is minimum.
i.e. 3 bags of brand P and 6 bags of brand Q; Minimum cost of the mixture = Rs 1950 .

Since, we know that

A−1=adjA|A|

pre multiply by A,

AA−1=AadjA|A|

I=AadjA|A|⇒AadjA=|A|I      (since AA-1=I)

On solving the equations  8x+4y=1 and 4x+8y=3, we get the point of intersection as (-1/2,5/12)

On solving the equations 8x+4y=5 and 4x+8y=7, we get the point of intersection as (1/4,3/4)

On solving the equations 8x+4y=1 and 4x+8y=7, weget the point of intersection as (-5/12,13/12)

On solving the equations 8x+4y=5and 4x+8y=3, we the point of intersection as (7/12,1/12)

Let the points A(1/12,5/12), B(7/12,1/12) C(1/4,3/4) and D(-5/12,13/12) be the vertices of the quadrilaeral

Since the slopes of the opposite sides are equal the quadrilateral is a parallelogram

Slope of the diagonal AC is 5/12−3/4−1/2−1/4 = 1

Slope of the diagonal BD is 1/2−13/127/12−(−5/12) = -1

Since the product of the slopes is -1, the diagonals are perpendicular to each other

Hence the parallelogram is a rhombus