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Linear programming — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsLinear programming1 Mark
MCQ
The objective function $Z = 4x + 3y$ can be maximised subjected to the constraints $3x + 4y ≤ 24, 8x + 6y ≤ 48, x ≤ 5, y ≤ 6, x, y ≥ 0$
  • A
    at only one point
  • B
    at two points only
  • ✓
    at an infinite number of points
  • D
    none of these
✓

Answer

Correct option: C.
at an infinite number of points
We need to maximize $Z = 4x + 3y$
First, we will convert the given inequations into equations, we obtain the following equations:
$3x + 4y = 24, 8x + 6y = 48, x = 5, y = 6, x = 0$ and $y = 0$.
The line $3x + 4y = 24$ meets the coordinate axis at $A(8, 0)$ and $B(0, 6).$
Join these points to obtain the line $3x + 4y = 24.$
Clearly, $(0, 0)$ satisfies the inequation $3x + 4y ≤ 24.$
So, the region in $xy-$plane that contains the origin represents the solution set of the given equation.
The line $8x + 6y = 48$ meets the coordinate axis at $C(6, 0)$ and $D(0, 8).$
Join these points to obtain the line $8x + 6y = 48.$
Clearly, $(0, 0)$ satisfies the inequation $8x + 6y ≤ 48.$
So, the region in $xy$ plane that contains the origin represents the solution set of the given equation.
$x = 5$ is the line passing through $x = 5$ parallel to the $Y$ axis.
$y = 6$ is the line passing through $y = 6$ parallel to the $X$ axis.
Region represented by $x ≥ 0$ and $y ≥ 0:$
Since, every point in the first quadrant satisfies these inequations.
So, the first quadrant is the region represented by the inequations.
These lines are drawn using a suitable scale.
and $B (0,6).$
The corner points of the feasible region are $O(0, 0), G(5, 0), \text{F}\Big(5,\frac{4}{3}\Big),\text{E}\Big(\frac{24}{7},\frac{24}{7}\Big)$and $B(0, 6).$
The values of $Z$ at these corner points are as follows.
$\text{Corner point}$ $\text{Z} = 4\text{x} + 3\text{y}$
$\text{O}(0, 0)$ $4 \times 0 + 3 \times 0= 0$
$\text{G}(5, 0)$ $4 \times 5 + 3 \times 0 = 20$
$\text{F}\Big(5,\frac{4}{3}\Big)$ $4\times5+3\times\frac{4}{3}=24$
$\text{E}\Big(\frac{24}{7},\frac{24}{7}\Big)$ $4\times\frac{24}{7}+3\times\frac{24}{7}=\frac{196}{7}=24$
$\text{B}(0, 6)$ $4\times0+3\times6=18$
We see that the maximum value of the objective function $Z$ is $24$ which is at $F(5, 4)$ and $\text{E}\Big(\frac{24}{7},\frac{24}{7}\Big).$
Thus, the optimal value of $Z$ is $24.$
As, we know that if a $\text{LPP}$ has two optimal solution, then there are an infinite number of optimal solutions.
Therefore, the given objective function can be subjected at an infinite number of points.

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