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Previous Year Question (PYQs)



The maximum value of $z=4x+2y$ subject to constraints
$2x+3y \le 28$,
$x+y \le 10$,
$x,y \ge 0$ is:





Solution

Corner points of feasible region:

$(0,0)$

From $x+y=10$
If $x=0$, $y=10$ → but check first constraint:
$2(0)+3(10)=30>28$ ❌ not feasible

So take intersection of

$2x+3y=28$
$x+y=10$

Solve:

$y=10-x$

Substitute:

$2x+3(10-x)=28$

$2x+30-3x=28$

$-x=-2$

$x=2$

Then

$y=10-2=8$

So intersection point $(2,8)$

Now check intercepts:

If $y=0$

From $x+y=10$ → $x=10$

Check first constraint:

$2(10)+0=20\le 28$ ✔

So point $(10,0)$ is feasible.

Now evaluate $z=4x+2y$

At $(0,0)$ → $0$
At $(2,8)$ → $4(2)+2(8)=8+16=24$
At $(10,0)$ → $40$

Maximum value = 40


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