Refresher - MATHEMATICS Quiz 23
PROBLEM 1:
One of the roots of the quadratic equation Ax2 + 14x + 12 = 0 is 6 times
the other root. Find the value of A.
Solution:
r2 = 6 r1
r1 r2 = C
A
r1 6 r1 = 12
A
r12 = 2
A
r1 + r2 = - B
A
r1 + 6 r1 = - 14
A
7 r1 = - 14
A
r1 = - 2
A
Square of r1 :
r1 2 = ( -A2 )2 = - 4
A2
r12 =2
A
2 = - 14
A A
A=2
, Refresher - MATHEMATICS Quiz 23
PROBLEM 2:
If P = 2x + 3y, find the maximum value of P subject to the following constraints:
⎧ x +y≤4
⎪⎪
2x + y ≤ 6 y
⎨
⎪ x ≥0
⎪⎩ y ≥ 0
2x + y = 6
(0,4)
(2,2)
R x+y=4
x
(0,0) (3,0)
Solution:
We solve the system of inequalities to find the feasibility region R shaded section
in the figure. The coordinates of its corner points are (0, 0), (3, 0), (0, 4), and
(2, 2).
Since the maximum value of P will occur at a corner of R, we substitute the
coordinates of each corner point into the objective function P = 2x + 3y and find
the one that gives the maximum value of P.
Point P = 2x + 3y
(0, 0) P = 2(0) + 3(0) = 0
(3, 0) P = 2(3) + 3(0) = 6
(2, 2)P = 2(2) + 3(2) = 10
(0, 4)P = 2(0) + 3(4) = 12
The maximum value P = 12 occurs when x = 0 and y = 4.
PROBLEM 1:
One of the roots of the quadratic equation Ax2 + 14x + 12 = 0 is 6 times
the other root. Find the value of A.
Solution:
r2 = 6 r1
r1 r2 = C
A
r1 6 r1 = 12
A
r12 = 2
A
r1 + r2 = - B
A
r1 + 6 r1 = - 14
A
7 r1 = - 14
A
r1 = - 2
A
Square of r1 :
r1 2 = ( -A2 )2 = - 4
A2
r12 =2
A
2 = - 14
A A
A=2
, Refresher - MATHEMATICS Quiz 23
PROBLEM 2:
If P = 2x + 3y, find the maximum value of P subject to the following constraints:
⎧ x +y≤4
⎪⎪
2x + y ≤ 6 y
⎨
⎪ x ≥0
⎪⎩ y ≥ 0
2x + y = 6
(0,4)
(2,2)
R x+y=4
x
(0,0) (3,0)
Solution:
We solve the system of inequalities to find the feasibility region R shaded section
in the figure. The coordinates of its corner points are (0, 0), (3, 0), (0, 4), and
(2, 2).
Since the maximum value of P will occur at a corner of R, we substitute the
coordinates of each corner point into the objective function P = 2x + 3y and find
the one that gives the maximum value of P.
Point P = 2x + 3y
(0, 0) P = 2(0) + 3(0) = 0
(3, 0) P = 2(3) + 3(0) = 6
(2, 2)P = 2(2) + 3(2) = 10
(0, 4)P = 2(0) + 3(4) = 12
The maximum value P = 12 occurs when x = 0 and y = 4.