Q.The maximum value of the objective function $Z=3x+5y$ subject to the constraints $x\geq 0$, $y\geq 0$ and $2x+5y\leq 10$ is :

Find the corner points of the feasible region and evaluate $Z$; the maximum is $15$ at $(5,0)$.

Step 1 — Feasible region. The constraints are $x\ge0,\ y\ge0,\ 2x+5y\le10$. The line $2x+5y=10$ meets the axes at:

• $y=0\Rightarrow x=5$, i.e. $(5,0)$
• $x=0\Rightarrow y=2$, i.e. $(0,2)$

So the feasible region is the triangle with vertices $(0,0),\ (5,0),\ (0,2)$.

Step 2 — Evaluate $Z=3x+5y$ at each corner.

The corner points give $Z=0,15,10$; the maximum is $15$ at $(5,0)$, so the answer is option (d).

We maximise $Z = 3x + 5y$ subject to

$2x + 5y \le 10,\quad x \ge 0,\quad y \ge 0.$

Find the corner points of the feasible region. The boundary line $2x + 5y = 10$ meets the axes at:

• $x$-axis $(y=0)$: $2x = 10 \Rightarrow x = 5$, point $(5, 0)$.
• $y$-axis $(x=0)$: $5y = 10 \Rightarrow y = 2$, point $(0, 2)$. …