Question

Maximize f(x) = 22x_{1} + 25x_{2} subject to the constraints 2x_{1} + x_{2} = 0, x_{2} >= 0​

Answer 1

Answer:

Z=3x+2y -------.. (i)

subject to the constraints

x+2y≤10

3x+y≤15

Convert these inequalities into equations

x+2y=10 .------. (ii)

3x+y=15 .------(iii)

From (ii), we get

x=0⟹y=5 and y=0 when x=10

So, the points (0,5) and (10,0) lie on the line given in (ii)

From (iii), we get the points

(0,15) and (5,0)

Let's plot these point and we get the graph in which, shaded part shows the feasible region.

(b)

Lines (ii) and (iii) intersect at (4,3) and other corner points of the region are (0,5),(5,0) and (0,0).

(c)

To find the maximum value of z, we need to find the value of z at the corner points

Corner points z=3x+2y

(0,0) 0

(5,0) 15

(0,5) 10

(4,3) 18

Thus, z is maximum at (4,3) and its maximum value is 18.

Answer 2

Answer:

We have, z=10x+25y .... (i)

The constraints are x≤3 ....(ii)

y≤3 .... (iii)

x+y≤5 .... (iv)

x≥0 .... (v)

y≥0 ....(vi)

The constraints (ii) and (iii) represent the closed half-planes on the left of the line x=3 and below y=3 respectively.

The line corresponding to (iv) is

x+y=5 ...(vii)

(5,0) and (0,5) lie on (vii). The origin does not lie on (vii) and it lies in half-plane of (vii) of 0+0≤5, which is true.

∴ The closed half plane containing the origin in the graph of (iv).

The constraints (v) and (vi) represent the closed half-planes on the right of y-axis and above x-axis respectively.

The shaded bounded region is the feasible region of the given L.P.P. we use curve point method to maximize z. The vertices of the feasible region are O(0,0),A[0,3),B(2,3),C(3,2) and D(3,0)

At O(0,0) z=10(0)+25(0)=0

At A(0,3) z=10(0)+25(3)=75

At B(2,3) z=10(2)+25(3)=95

At C(3,2) z=10(3)+25(2)=80

At D(3,0) z=10(3)+25(0)=30

∴ maximum value of z=95, where x=2 and y=3.