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MHT CET · Maths · Area Under Curves

The objective function Z=4x1+5x2, subject to 2x1+x27, 2x1+3x215,x23,x1,x20 has minimum value at the point

  1. A On x axis
  2. B On y axis
  3. C At the origin
  4. D On the line parallel to x axis
Verified Solution

Answer & Solution

Correct Answer

(A) On x axis

Step-by-step Solution

Detailed explanation

Value of \(\mathrm{z}=4 \mathrm{x}_{1}+5 \mathrm{x}_{2}\)
Convert the given inequalities into equalities to get the corner points \(2 x_{1}+x_{2}=7 \ldots \ldots\) (i)
At \(x_{1}=0, x_{2}=7\) and \(x_{2}=0, x_{1}=3.5\)
So, the corner points of (i) are \((0,7)\) and \((3.5,0)\)
\(2 \mathrm{x}_{1}+3 \mathrm{x}_{2}=15 \ldots \ldots\) (ii)
At \(\mathrm{x}_{1}=0, \mathrm{x}_{2}=5\) and \(\mathrm{x}_{2}=0, \mathrm{x}_{1}=7.5\)
So, the corner points of (i) are \((0,5)\) and \((7.5,0)\)
\(\mathrm{x}_{2}=3\)..... (iii)
Plot these corner points on the graph paper and the line given in (iii)
The shaded part shows the feasible region.
At \(x_{2}=3, x_{1}=2\) in (i) and \(x_{1}=3\) in (ii)
The corner points of the feasible region are \((3.5,0),(7.5,0),(3,3)\) and \((2,3)\)
Corner points \(Z=4 x_{2}+5 x_{2}\)
\(\begin{array}{ll}(3.5,0) & 14 \\ (7.5,0) & 30 \\ (3,3) & 27 \\ (2,3) & 23\end{array}\)
Minimum value of \(Z=14\) and it lies on \(x\)-axis.