Q.Find the value of the following: Maximise $Z = 5x + 3y$ subject to $3x + 5y \le 15$, $5x + 2y \le 10$, $x \ge 0$, $y \ge 0$.

The key idea is that the maximum of a linear objective under linear constraints occurs at a corner point of the feasible region.

Step 1: Find the corner points from the constraints.

Solve $3x + 5y = 15$ and $5x + 2y = 10$:

Multiply first by 2, second by 5: $6x + 10y = 30$, $25x + 10y = 50$. Subtract: $19x = 20 \Rightarrow x = \frac{20}{19}$, then $y = \frac{15 - 3(20/19)}{5} = \frac{225}{95} = \frac{45}{19}$.

Other corners: $(0,0)$, $(2,0)$ from $5x+2y=10$ with $y=0$, and $(0,3)$ from $3x+5y=15$ with $x=0$.

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

• $(0,0)$: $Z = 0$
• $(2,0)$: $Z = 10$
• $(0,3)$: $Z = 9$
• $\left(\frac{20}{19}, \frac{45}{19}\right)$: $Z = 5\cdot\frac{20}{19} + 3\cdot\frac{45}{19} = \frac{100 + 135}{19} = \frac{235}{19} \approx 12.37$

Step 3: The largest value is $\frac{235}{19}$ at the intersection point.

We are asked to maximise a linear objective function $Z = 5x + 3y$ subject to two linear inequalities and the usual non-negativity conditions. This is a classic linear programming problem in two variables — the kind you solve graphically by identifying the feasible region and checking its corner points.

The key idea: the maximum (or minimum) of a linear function over a convex polygon (the feasible region) always occurs at a vertex, or along an entire edge if the objective is parallel to it. So we don't need to test every point — just the corners.

Let’s work through it.

1. Write down the constraints clearly

2. Find the boundary lines

   For Constraint 1: $3x + 5y = 15$

   • If $x = 0$, then $y = 3$ → point $(0,3)$
   • If $y = 0$, then $x = 5$ → point $(5,0)$

   For Constraint 2: $5x + 2y = 10$

   • If $x = 0$, then $y = 5$ → point $(0,5)$
   • If $y = 0$, then $x = 2$ → point $(2,0)$

3. Identify the feasible region

   Since both constraints are “$\le$” and $x, y \ge 0$, the feasible region is the set of points in the first quadrant that lie below both lines.

   The region is a polygon with vertices at:

   • $(0,0)$ — origin
   • $(2,0)$ — where Constraint 2 meets the $x$-axis
   • $(0,3)$ — where Constraint 1 meets the $y$-axis
   • The intersection point of the two lines (if it lies in the first quadrant)

4. Find the intersection of the two lines

   Solve:

Multiply the first equation by 2 and the second by 5 to eliminate $y$:

Subtract: $(25x - 6x) + (10y - 10y) = 50 - 30$

Substitute into $3x + 5y = 15$:

So the intersection point is $\left(\frac{20}{19}, \frac{45}{19}\right)$.

5. List all corner points

   The feasible region has four vertices:

   • $A(0,0)$
   • $B(2,0)$
   • $C\left(\frac{20}{19}, \frac{45}{19}\right)$
   • $D(0,3)$

   (The point $(0,5)$ is not feasible because it violates Constraint 1; $(5,0)$ violates Constraint 2.)

6. Evaluate $Z = 5x + 3y$ at each corner

   • At $A(0,0)$: $Z = 5(0) + 3(0) = 0$
   • At $B(2,0)$: $Z = 5(2) + 3(0) = 10$
   • At $C\left(\frac{20}{19}, \frac{45}{19}\right)$:

• At $D(0,3)$: $Z = 5(0) + 3(3) = 9$

7. Compare values

The largest is $\frac{235}{19}$ at point $C$.

Method: Solving a two-variable LPP graphically (maximisation with an interior vertex)

The standard graphical method, with emphasis on the vertex where two constraints are simultaneously active — often the point that carries the optimum.

Steps

Step 1: Plot the constraint lines and locate the feasible region.

Draw each line by its intercepts, shade using an origin test, and intersect with the first quadrant ($x,y\ge0$).

Step 2: Find every genuine corner, including the two-line intersection.

Solve the two slanted constraints simultaneously (eliminate one variable). This interior vertex is easy to forget, yet the maximum frequently lives there — not at an axis intercept.

Step 3: Evaluate and maximise.

Compute $Z$ at each valid corner and pick the largest. The optimum can be a fraction; carry exact values rather than rounding, since board marking rewards the exact corner value.

Common Mistakes

Mistake 1: Skipping the two-line intersection vertex.

Why it's wrong: the maximum here is at $\left(\tfrac{20}{19},\tfrac{45}{19}\right)$, where both constraints are active — checking only the intercepts $(2,0)$ and $(0,3)$ misses it. Correct approach: always solve the constraints simultaneously to get the interior vertex.

Mistake 2: Rounding the fractional optimum too early.

Why it's wrong: the exact maximum is $\tfrac{235}{19}$; premature rounding loses marks and can mis-rank corners. Correct approach: keep exact fractions when comparing corner values.