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Maximise 3x + 4y subject to the constraints x + y ≤ 5, x ≤ 3, y ≤ 4, x ≥ 0, y ≥ 0.

To solve this linear programming problem using graphical method, we first plot the feasible region by graphing the constraints.

The constraint x + y ≤ 5 can be graphed as a straight line with intercepts (0,5) and (5,0). The constraint x ≤ 3 can be graphed as a vertical line passing through (3,0). The constraint y ≤ 4 can be graphed as a horizontal line passing through (0,4).

We then shade the feasible region, which is the region that satisfies all the constraints. In this case, the feasible region is the triangle bounded by the lines x + y = 5, x = 3, and y = 4.

Next, we plot the objective function 3x + 4y = k on the same graph. We can do this by choosing a value of k and graphing the line 3x + 4y = k. We then move the line parallel to itself until it is tangent to the feasible region at a corner point.

We repeat this process for different values of k until we find the maximum value of 3x + 4y. In this case, the maximum value occurs at the corner point (3,2) and is equal to 3(3) + 4(2) = 17.

Therefore, the solution to the linear programming problem is x = 3, y = 2, and the maximum value of the objective function is 17.

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