Mathematics - Online Test
Let number of packages of nuts produced = x
And number of packages of bolts produced = y
Therefore , the above L.P.P. is given as :
Maximise , Z = 17.50x +7y , subject to the constraints : x +3y ≤ 12 and. 3x +y ≤ 12, x, y ≥ 0.
Corner points | Z =17.50 x +7 y |
O( 0 , 0 ) | 0 |
D(4,0 ) | 70 |
A(0,4) | 28 |
B(3,3) | 73.50…………………(Max.) |
Here Z = 73.50 is maximum.
i.e 3 packages of nuts and 3 packages of bolts;
Maximum profit = Rs 73.50.
Tangents to the curve at the points (1, 1) and ( – 1, 1)
therefore , slope of tangent at (1,1) = - 1 and the slope of tangent at ( - 1 ,1 )= 1 .
Now product of the slopes=1.-1= -1
Hence , the two tangents are at right angles.
The lines x+1=0 and y+1=0 are perpendicular to each other.
The slope of the line x+y =0 is -1
Hence the angle made by this line with respect to X axis is 450
In other words the angle made by this line with x+1=0 is 450
Clearly the other line with which it can make 450 is y+1=0
p: x = 5 and y = - 2 , q : x – 2y = 9
The contrapositive of
Hence If x – 2y 9 , then x 5 or y - 2
