Question

Solve the inequality $3x + 4y \leq 12$, $4x + 3y \leq 12$, $x \geq 0$, $y \geq 0$ by graphical method.

Hint:

In graphical inequalities → always test intercepts, plot boundary lines, and then shade the common feasible region.

Answer 1

We have the system of inequalities: \[ 3x + 4y \leq 12, 4x + 3y \leq 12, x \geq 0, y \geq 0 \]

Step 1: Convert inequalities into equations. \[ 3x + 4y = 12 (1) \] \[ 4x + 3y = 12 (2) \]

Step 2: Find intercepts. For (1): - If $x=0 \implies y = 3$ - If $y=0 \implies x = 4$ For (2): - If $x=0 \implies y = 4$ - If $y=0 \implies x = 3$

Step 3: Plot lines and shade feasible region. - Line (1): passes through $(0,3)$ and $(4,0)$. - Line (2): passes through $(0,4)$ and $(3,0)$. - Feasible region lies in the first quadrant bounded by these lines and coordinate axes.

Step 4: Find corner points of feasible region. By solving (1) and (2): \[ 3x + 4y = 12, 4x + 3y = 12 \] Multiply first by 3: $9x + 12y = 36$ Multiply second by 4: $16x + 12y = 48$ Subtract: $7x = 12 \implies x = \dfrac{12}{7}$ Substitute in (1): \[ 3\left(\dfrac{12}{7}\right) + 4y = 12 \implies \dfrac{36}{7} + 4y = 12 \] \[ 4y = \dfrac{84 - 36}{7} = \dfrac{48}{7} \implies y = \dfrac{12}{7} \] So, point of intersection is $\left(\dfrac{12}{7}, \dfrac{12}{7}\right)$.

Step 5: Vertices of feasible region. \[ (0,0), \; (4,0), \; (0,4), \; \left(\dfrac{12}{7}, \dfrac{12}{7}\right) \] Conclusion: The shaded feasible region is the quadrilateral formed by these points in the first quadrant.

Final Answer: Feasible region vertices are \[ \boxed{(0,0), \; (4,0), \; (0,4), \; \left(\dfrac{12}{7}, \dfrac{12}{7}\right)} \]