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Linear Programming — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsLinear Programming5 Marks
Question
We first draw the lines processed on three machines : Assembling, Finishing and Polishing. The time required for each product in hours and availability of each machine is given by following table:

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Formulate the above problem as L.P.P. Solve it graphically to get maximum profit.

✓

Answer

Let the number of chairs and tables made by the carpenter be x and y respectively. The profits are ₹ 140 per chair and ₹ 210 per table. ∴ total profit z = ₹ (140x + 210y) This is the objective function which is to be maximized. The constraints are as per the following table :

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From the table, the constraints are

$3 x+3 y \leq 36,5 x+2 y \leq 50,2 x+6 y \leq 60$

The number of chairs and tables cannot be negative. ∴ x ≥ 0, y ≥ 0 Hence, the mathematical formulation of given LPP is : Maximize z = 140x + 210y, subject to

$3 x+3 y \leq 36,5 x+2 y \leq 50,2 x+6 y \leq 60, x \geq 0, y \geq 0$

We first draw the lines AB, CD and EF whose equations are 3x + 3y = 36, 5x + 2y = 50 and 2x + 6y = 60 respectively.

Image

The feasible region is OCPQFO which is shaded in the graph. The vertices of the feasible region are O (0, 0), C (10, 0), P, Q and F (0, 10). P is the point of intersection of the lines 5x + 2y = 50 … (1) and 3x + 3y = 36 … (2) Multiplying equation (1) by 3 and equation (2) by 2, we get 15x + 6y = 150 6x + 6y = 72 On subtracting, we get 26

$9 x=78 \therefore x=\frac{26}{3}$

Substituting $x=\frac{26}{3}$ in (2), we get

$3\left(\frac{26}{3}\right)+3 y=36$

$3 y=10 y=\frac{10}{3}$

Q is the point of intersection of the lines 3x + 3y = 36 … (2) and 2x + 6y = 60 … (3) Multiplying equation (2) by 2, we get 6x + 6 y = 72 Subtracting equation (3) from this equation, we get 4x = 12 ∴ x = 3 Substituting x = 3 in (2), we get 3(3) + 3y = 36 ∴ 3y = 27 ∴ y = 9 ∴ Q is (3, 9). Hence, the vertices of the feasible region are O (0, 0),

$C(10,0), \mathrm{P}\left(\frac{26}{3}, \frac{10}{3}\right), \mathrm{Q}(3,9)$ and $\mathrm{F}(0,10)$.

The values of the objective function z = 140x + 210y at these vertices are z(O) = 140(0) + 210(0) = 0 + 0 = 0 z(C) = 140 (10) + 210(0) = 1400 + 0 = 1400

$z(\mathrm{P})=140\left(\frac{26}{3}\right)+210\left(\frac{10}{3}\right)=\frac{3640+2100}{3}=\frac{5740}{3}=1913.33$

z(Q) = 140(3) + 210(9) = 420 + 1890 = 2310 z(F) = 140(0) + 210(10) = 0 + 2100 = 2100 ∴ z has maximum value 2310 when x = 3 and y = 9 Hence, the carpenter should make 3 chairs and 9 tables to get the maximum profit of ₹ 2310.

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