Answer:
Let David invests Rs. x in bond A and Rs. y in bond B. The given information can be written in tabular form as
Then, required linear programming problem is Maximize interest \[Z=8%\]of \[x+10%\]of y \[=0.08x+0.10y\] Subject to the constraints \[x+y\le 12000,\] \[x\ge 2000,\] \[y\ge 4000\] and \[x\ge 0,\,\,y\ge 0\] Consider the constraints as equations, we get \[x+y=12000\] ...(i) \[x=2000\] ...(ii) \[y=4000\] ...(iii) and \[x,\text{ }y=0\] Table for \[x+y=12000\] is Bonds Interest rates Restrictions A(x) 8% At least Rs. 2000 B(y) 10% At least Rs. 4000
So, line passes through the points (0, 12000) and (12000, 0). Putting (0, 0) in the inequality \[x+y\le 12000,\] we get \[0+0\le 12000\] \[\Rightarrow \] \[0\le 12000\] [true] \[\therefore \] The shaded region is towards the origin. \[\because \] Line x = 2000 is parallel to Y-axis. Putting (1000, 0) in the inequality \[x\ge 2000,\] we get \[1000\ge 2000\] [false] \[\therefore \] The shaded region is at the right side of the line. \[\because \] Line y = 4000 is parallel to X-axis. Putting (0, 6000) in the inequality \[y\ge 4000,\] we get \[6000\ge 4000\] [true] \[\therefore \] The shaded region is above the line. The intersection point of lines (ii) and (iii), (i) and (iii), (i) and (ii) are respectively, A (2000, 4000), B (8000, 4000) and C(2000,10000) Now, plot the graph of the system of inequalities. The shaded portion ABC represents the feasible region which is bounded. And the coordinates of the corner points are A(2000, 4000), 6(8000, 4000) and C(2000,10000), respectively Now, the values of Z at each comer point are given below x 0 12000 y 12000 0
\[\therefore \] Maximum value of Z is 1160 at (2000, 10000). Hence, maximum profit is Rs.1160 when Rs. 2000 are invested in bond A and Rs. 10000 are invested in bond B. Value The growth of money is also important to fulfil basic needs of life and investing can help a person achieve long life goal easily Corner points \[\mathbf{Z=0}\mathbf{.08x+0}\mathbf{.10y}\] A(2000, 4000) \[Z=0.08(2000)+0.10(4000)\] \[=160+400=560\] B(8000, 4000) \[Z=0.08(8000)+0.10(4000)\]\[=640+400=1040\] C(2000, 10000) \[Z=0.08(2000)+0.10(10000)\]\[=160+1000=1160\](maximum)
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