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Q.The feasible region for an LPP is shown in the graph given below : The graph shows the constraint line CDCD through C(0,6)C(0, 6) and D(12,0)D(12, 0), the line EFEF through E(0,4)E(0, 4) and F(5,0)F(5, 0), and the line through A(0,12)A(0, 12) and D(12,0)D(12, 0); the shaded feasible region is the quadrilateral with vertices C(0,6)C(0, 6), the interior intersection XX, BB (on the x-axis near x=6x=6) and E(0,4)E(0, 4). Based on the above information, answer the following questions :

(i) Determine the equation of CD.
(ii) Determine the equation EF.
(iii)
(a) Determine all the constraints for the LPP.
(OR)
(iii)
(b) Find the maximum value of the objective function Z=600x+400yZ = 600x + 400y.
CBSECBSE Class XII Board 2025Subjective· 4mImportance★★★★★
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CDCD through (0,6),(12,0)(0,6),(12,0) is x+2y=12x+2y=12; EFEF through (0,4),(5,0)(0,4),(5,0) is 4x+5y=204x+5y=20; with the third line AB: 2x+y=12AB:\ 2x+y=12, the constraints are x+2y≤12, 2x+y≤12, 4x+5y≥20, x,y≥0x+2y\le12,\ 2x+y\le12,\ 4x+5y\ge20,\ x,y\ge0; ZZ is maximum 40004000 at (4,4)(4,4).

Two-point / intercept form of a line through (a,0)(a,0) and (0,b)(0,b): xa+yb=1\dfrac{x}{a}+\dfrac{y}{b}=1. In an LPP the objective attains its optimum at a corner of the feasible region.

(i) Equation of CDCD

  1. CDCD passes through C(0,6)C(0,6) and D(12,0)D(12,0), so xx-intercept =12=12, yy-intercept =6=6.
  2. Intercept form: x12+y6=1\dfrac{x}{12}+\dfrac{y}{6}=1.
  3. Multiply by 1212: x+2y=12x+2y=12.

(ii) Equation of EFEF

  1. EFEF passes through E(0,4)E(0,4) and F(5,0)F(5,0), so xx-intercept =5=5, yy-intercept =4=4.
  2. Intercept form: x5+y4=1\dfrac{x}{5}+\dfrac{y}{4}=1.
  3. Multiply by 2020: 4x+5y=204x+5y=20.

(iii)(a) All the constraints

  1. The third boundary line runs through A(0,12)A(0,12) and B(6,0)B(6,0): intercept form x6+y12=1⇒2x+y=12\dfrac{x}{6}+\dfrac{y}{12}=1\Rightarrow 2x+y=12.
  2. Test the feasible region (which lies on the origin side of CDCD and ABAB but away from the origin relative to EFEF) using an interior point such as X(4,4)X(4,4):
  • x+2y=4+8=12≤12x+2y=4+8=12\le12 ⇒\Rightarrow x+2y≤12x+2y\le12.
  • 2x+y=8+4=12≤122x+y=8+4=12\le12 ⇒\Rightarrow 2x+y≤122x+y\le12.
  • 4x+5y=16+20=36≥204x+5y=16+20=36\ge20 ⇒\Rightarrow 4x+5y≥204x+5y\ge20 (origin gives 0≥200\ge20, false, so region is the far side). …

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