Find the maximum value of $Z = 3 x + 4 y$ subject to $2 x + y l e 10$ , $x + 3 y l e 12$ , $xg e 0$ , $y g e 0$ .

hard 5 marks Class 11 · Mathematics · Linear Inequalities

Answer

22

The correct answer is $22$ — The largest value is $22$ attained at $(x, y) = (3.6, 2.8)$ .

Step-by-step solution

List the four constraints and note the feasible region is a convex polygon.
Find intersection points of each pair of boundary lines: $(0, 0)$ , $(5, 0)$ from $2 x + y = 10$ , $(0, 4)$ from $x + 3 y = 12$ .
Solve $2 x + y = 10$ and $x + 3 y = 12$ simultaneously: $y = 10 - 2 x$ , substitute to get $x + 3 (10 - 2 x) = 12$ , yielding $x = 18/5$ , $y = 14/5$ (i.e. $3.6, 2.8$ ).
Check each vertex satisfies all inequalities.
Evaluate $Z$ at each vertex: $(0, 0) : 0$ , $(5, 0) : 15$ , $(0, 4) : 16$ , $(3.6, 2.8) : 22$ .
The largest value is $22$ attained at $(x, y) = (3.6, 2.8)$ .

Common mistake: Ignoring feasibility of intersection points or miscalculating coordinates

Concept tested: Linear programming optimisation