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NCERT Exemplar · Q10

Q.The feasible region of a linear programming problem is a bounded quadrilateral with corner points P(313,2413)P\left(\dfrac{3}{13}, \dfrac{24}{13}\right), Q(32,154)Q\left(\dfrac{3}{2}, \dfrac{15}{4}\right), R(72,34)R\left(\dfrac{7}{2}, \dfrac{3}{4}\right) and S(187,27)S\left(\dfrac{18}{7}, \dfrac{2}{7}\right). Determine the maximum and minimum values of Z=x+2yZ = x + 2y over this region.

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On the bounded quadrilateral the linear objective Z=x+2yZ = x + 2y takes corner values 5113≈3.92\frac{51}{13}\approx 3.92, 99, 55 and 227≈3.14\frac{22}{7}\approx 3.14 at P,Q,R,SP,Q,R,S. The maximum is 99 at QQ and the minimum is 227\frac{22}{7} at SS.

Concept

By the Corner Point Theorem, a linear objective on a bounded region attains its maximum and minimum at vertices. We evaluate Z=x+2yZ = x + 2y at each corner.

Evaluate Z=x+2yZ = x + 2y

Z(P)=313+2⋅2413=3+4813=5113≈3.92,Z(P)=\frac{3}{13}+2\cdot\frac{24}{13}=\frac{3+48}{13}=\frac{51}{13}\approx 3.92,

Z(Q)=32+2⋅154=32+152=182=9,Z(Q)=\frac{3}{2}+2\cdot\frac{15}{4}=\frac{3}{2}+\frac{15}{2}=\frac{18}{2}=9,

Z(R)=72+2⋅34=72+32=102=5,Z(R)=\frac{7}{2}+2\cdot\frac{3}{4}=\frac{7}{2}+\frac{3}{2}=\frac{10}{2}=5,

Z(S)=187+2⋅27=18+47=227≈3.14.Z(S)=\frac{18}{7}+2\cdot\frac{2}{7}=\frac{18+4}{7}=\frac{22}{7}\approx 3.14. …

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