Linear Programming

Let the daily production be $x$ necklaces and $y$ bracelets.

Formulation

Maximise profit $Z = 80x + 50y$ subject to

• $x + y \le 24$ (handling capacity)
• $\tfrac12 x + y \le 16$, i.e. $x + 2y \le 32$ (labour hours)
• $x \ge 0,\ y \ge 0$

Graph

Draw the line $x + y = 24$ through $(24, 0)$ and $(0, 24)$, and the line $x + 2y = 32$ through $(32, 0)$ and $(0, 16)$. The feasible region is the area on the origin side of both lines in the first quadrant. The lines meet where $x + y = 24$ and $x + 2y = 32$, giving $y = 8$, $x = 16$.

The corner points of the feasible region are $O(0,0)$, $A(24,0)$, $B(16,8)$ and $C(0,16)$.

The maximum profit is at $A(24, 0)$.

Conclusion: Produce 24 necklaces and no bracelets daily for a maximum profit of Rs. 1920.

(b) a convex polygon

The feasible region of a linear programming problem is the intersection of half-planes given by linear constraints, and the intersection of convex sets is always a convex set — so it is a convex polygon (bounded or unbounded).

The feasible region is bounded by $x+y=6$, $x=4$, $y=3$, $x=0$ and $y=0$. Its corner points are found by intersecting these lines (note $x=4$ meets $x+y=6$ at $(4,2)$, and $y=3$ meets it at $(3,3)$).

The maximum value is $Z_{\max}=16$, occurring at $(x,y)=(4,2)$.

Let $x$ = number of units of A and $y$ = number of units of B produced daily.

Formulation

$\text{Maximize } Z=40x+30y$

Subject to $2x+y\le12 \quad\text{(labour hours)}$ $x+2y\le12 \quad\text{(machine hours)}$ $x\ge0,\; y\ge0$

Solution (corner-point method)

Corner points of the feasible region: $(0,0)$, $(6,0)$, $(0,6)$ and the intersection of $2x+y=12$ and $x+2y=12$.

From $y=12-2x$: $x+24-4x=12 \Rightarrow x=4,\; y=4$, giving $(4,4)$.

The maximum profit is Rs 280 per day, so the factory should produce 4 units of A and 4 units of B daily.

Formulation

Let $x$ and $y$ be the units of $F_1$ and $F_2$ in the meal.

$\text{Minimize } Z=5x+4y$

Subject to $3x+2y\ge30 \quad\text{(vitamin A)}$ $2x+y\ge20 \quad\text{(vitamin B)}$ $x+2y\ge12 \quad\text{(vitamin C)}$ $x\ge0,\; y\ge0$

Graphical solution

Intercepts of the boundary lines:

The intersection of $2x+y=20$ and $x+2y=12$ is $\left(\frac{28}{3},\frac43\right)$.

The vitamin A constraint is redundant: since $3x+2y=\frac43(2x+y)+\frac13(x+2y)\ge\frac43(20)+\frac13(12)=\frac{92}{3}>30$, any point satisfying the other two also satisfies it. The unbounded feasible region therefore has corner points $A(0,20)$, $B\left(\frac{28}{3},\frac43\right)$ and $C(12,0)$.

The region is unbounded, but $Z$ has positive coefficients, so its minimum lies at a corner point.

Conclusion: mix $\frac{28}{3}$ units of $F_1$ with $\frac43$ units of $F_2$; the minimum cost is Rs 52.