Solve the following linear programming problem graphically:

Maximise $Z = 34 x + 45 y$

under the following constraints

$x + y \leq 300 2 x + 3 y \leq 70 x \geq, y \geq 0$

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Solution

Here, objective function of given LPP is:

Maximise $Z = 34 x + 45 y$

Subject to the constraints:

$x + y \leq 300 2 x + 3 y \leq 70 x \geq, y \geq 0$

Consider $x + y = 300$

Table of solutions is:

$\begin{matrix} x & 300 & 0 y & 0 & 300 \end{matrix}$

Consider $2 x + 3 y = 70$

Table of solutions is:

$\begin{matrix} x & 35 & 2 y & 0 & 22 \end{matrix}$

To solve the LPP, we draw the graph of the inequations and get the feasible solution shown (shaded)

in the graph,. Corner points of the common shaded region are $P (0, 0), A (35, 0) a n d B (0, \frac{70}{3}) .$

Value of $Z$ at each corner point is given as
$\begin{matrix} Corner Points & Value of the objective function Z = 34 x + 45 y At O (0, 0) & Z = 0 + 0 = 0 At A (35, 0) & Z = 34 \times 35 + 0 = 1190 At B (0, \frac{70}{3}) & Z = 0 + 45 \times \frac{70}{3} = 1050 \end{matrix}$

Hence, maximum value of $Z$ is $1190$ and it is obtained when $x = 35$ and $y = 0$

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