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John has 20,000toinvestinthreefunds{F}_{1},{F}_{2}and{F}_{3}.Fund{F}_{1}isoffersareturnof2{F}_{2}offersareturnof4{F}_{3}offersareturnof5 3000 in F3 and at least twice as much as in F1 than in F2. Assuming that the rates hold till the end of the year, what amounts should he invest in each fund in order to maximize the year end return?

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Solution

Let x be the amount invested in F1, y the amount invested in F2 and z the amount invested in F1.
x+y+z=20,000
z=20,000(x+y)
Total return R of all three funds is given by
R=2%x+4%y+5%z=0.2x+0.04y+0.05(20,000(x+y))
Simplifies to
R(x,y)=10000.03yx0.01y: This is the return to maximize
Constraints: x,y0 which may be written as x+y20,000
John invests no more than 3000in{F}_{3}$, hence
z3000
Substitute z by 20,000(x+y) in the above inequality to obtain
20,000(x+y)3000 which may be written as x+y17,000
Let us put all the inequalities together to obtain the following system
⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪x0x0(x+y)17000x+y20,000x2y
The solution set of the system of inequalities above and the vertices of the feasible solution set obtained are shown.
Vertices
A at intersection of x+y=20000 and y=0, coordinates of A:(20000,0)
B at intersection of x+y=17000 and y=0 coordinates of B:(17000,0)
C at intersection of x+y=17000 and x=2y coordinates of C:(11333,5667)
D is at intersection of x=2y and x+y=20000, coordinates of D:(13333,6667)
Evaluete the return R(X,y)=10000.03x0.01y at each one of the vertices A(x,y),B(x,y),C(x,y) and D(x,y)
At A(20000,0):R(20000,0)=10000.03(20000)0.01(0)=400
At B(17000,0):R(17000,0)=10000.03(17000)0.01(0)=490
At C(11333,5667):R(11333,5667)=10000.03(13333)0.01(5667)=603
At D(13333,6667):R(13333,6667)=10000.03(13333)0.01(6667)=533
The return R is maximum at the vertex at C(11333,5667) where x=11333 and y=5667 and z=20,000(x+y)=3000
For maximum return, John has to invest 11333infund{F}_{1}, 5667 in fund F2 and 3000infund{F}_{3}$
1032509_849428_ans_7494e7ebe0db429db4375a3375d624be.PNG

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