An investment club has set a goal of earning 15% on the money they invest in sto

Question

An investment club has set a goal of earning 15% on the money they invest in stocks. The members are considering purchasing three possible stock5, with their cost per 5hare in dollars) and their project growth per share (in dollars) summarized in the lable. (Lel x conputer shares, y utility shares, and z relail shares.) Stocks Computer (x) Utility (y) Retail (z) Cost/share 30 26 Growth/share 6.00 6.00 2.40 (a) If they have $392,000 to invest, how many shares of each stock should they buy to meet their goal? (If there are infinitely many solutions, express your answers in terms of z as in Example 3.) (h) f they huy 600 shares of retail stock, how many shares of the other stocks do they buy? computer utility shares shares What If they buy 1200 shares of retall stock? computer utility shares shares (c) What is the minimum number of shares of computer stock they will buy? shares What is the number of shares of the other stocks in this case? utility retail shares shares (d) What is the maximum number ot shares of computer stock purchased? sharcs What is the number of shares of the other stocks in this casc? utility retail shares shares

Explanation / Answer

(a).Let the number of shares of stocks x,y,z to be purchased be a,b and c respectively. Then 30a +44b +26c = 392000 or, on dividing both the sides by 2, 15a+22b+13c = 196000 …(1). Further, growth per share in the stocks x,y,z being 6.00,6.00.2.40 respectively, we have6a+ 6b +2.40c= $ 392000*15/100 or, 60a+60b+24c = 588000 or, on dividing both the sides by 12, 5a+5b + 2c = 49000…(2)

The augmented matrix of this system of linear equations is A(say) =

15

22

13

196000

5

5

2

49000

To solve this system of linear equations, we will reduce A to its RREF as under:

Multiply the 1st row by 1/15

Add -5 times the 1st row to the 2nd row

Multiply the 2nd row by -3/7

Add -22/15 times the 2nd row to the 1st row

Then the RREF of A is

1

0

-3/5

2800

0

1

1

7000

This implies that a-3c/5=2800 or,a=2800+3c/5 and b+c=7000 or, b =7000-c.Then (a,b,c) =(2800+3c/5, 7000-c, c )=(2800,7000,0) +c/5(3,-5, 5) =(2800,7000,0) +t(3,-5, 5) = (2800+3t, 7000-5t, 5t), where t is an arbitrary positive real number. Now, since the number of shares of any stock cannot be negative,hence,7000-5t 0 or, t 1400. Thus, the solution is, the number of shares of stocks (x,y,z ) = (2800+3t, 7000-5t, 5t), where t 1400. There are , thus, infinite solutions.

(b). If 5t = 600, then t = 120 so that 2800 +3t = 2800 +360 = 3160 and 7000-5*120 = 7000-600 = 6400.

Thus, the no. of computerstock shares will be 3160 and

the no. of utility stock shares will be 6400.

(c ). The least value of t is 0 so that the minimum number of computerstock shares will be 2800+ 3*0 = 2800.

(d). The maximum value of t is 1400 so that the maximum number of computerstock shares will be 2800+3*1400 = 7000. Further, when t = 1400, then 7000-5t = 0 and 5t = 7000 so that, in this case, the number of utility stock shares will be 0 and

the no. of retail stock shares will be 7000.

15

22

13

196000

5

5

2

49000

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