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Linear programming — Maths STD 12 Science — Question

Gujarat 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 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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