Question

Problem 2-10 (Algorithmic) For the linear program Max 3 A + 3 B s.t. A + 3B ≤ 9 10A + 6B ≤ 30 A, B ≥ 0 select the correct graph that identifies the optimal solution. What is the value of the objective function at the optimal solution? (i) BA (ii) BA (iii) BA (iv) BA The value of the objective function at the optimal solution is .

Answer 1

According to the given question, the objective function is written as:

Max 3A +3B

subject to the constraint

$A+3B\leq 9.............................(a)$

10A6B< 30

and

A,B>O

By graphical method we can solve the linear program as follows;

In order to determined the straight line graph we consider two constraint as :

$A+3B=9$

divide by 9 both side we get

$\Rightarrow \frac{A}{9}+\frac{B}{3}=1............(i)$

Similarly

$\Rightarrow \frac{A}{3}+\frac{B}{5}= 1..................(ii)$

The graph of this two equation is shown as follows:

10 Constraint (i) A+3B-9 6 4 x(3,0) y(A 3/2,B-5/2) 2 Constrain (ii) 10A+6B-30 (0,0) 4 10 2 z(0,3)

As this two constraint is less than type , therefore the feasible zone of this constraint should be towards the origin and hence the feasible solution at point

$x(A=3,B=0)$ ,$z(A=0,B=3)$ and by solving equation (i) and (ii) we get $y(A=\frac{3}{2},B=\frac{5}{2})$ as

$10A+6B=30$

$2A+6B=18$

.(-)......(-)............(-)..............................

$8A=12$

$\Rightarrow A=\frac{12}{8}=\frac{3}{2}$

and hence

$\Rightarrow B=\frac{5}{2}$

Therefore the optimum solution at point x,y,z and origin is determined as:

$Max_{x}:3A+3B=3\times 3+3\times 0$

$Max_{x}\Rightarrow 9$

$Max_{y}:3A+3B=3\times \frac{3}{2}+3\times \frac{5}{2}$

$Max_{y}\Rightarrow 12$

$Max_{z}:3A+3B=3\times 0+3\times 3$

$Max_{z}\Rightarrow 9$​​​​​​​

$Max_{0}:3A+3B=3\times 0+3\times 0$

$Max_{o}\Rightarrow 0$​​​​​​​

Therefor the optimum solution at point y where
2 A=
and $B=\frac{5}{2}$ and maximum value of the objective function is determined as:

$Max_{y}:3A+3B=3\times \frac{3}{2}+3\times \frac{5}{2}$

$Max_{y}\Rightarrow 12$​​​​​​​​​​​​​​