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 stocks, with their cost per share (in dollars) and their projected growth per share (in dollars) summarized in the table. (Let x = computer shares, y = utility shares, and z = retail shares.)
Stocks
Computer (x)   Utility (y)   Retail (z)
Cost/share   30   44   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.)

b) If they buy 1500 shares of retail stock, how many shares of the other stocks do they buy?
computer

c)What if they buy 3000 shares of retail stock?
computer
utility

(c) What is the minimum number of shares of computer stock they will buy?

D)What is the number of shares of the other stocks in this case?
utility
retail

(d) What is the maximum number of shares of computer stock purchased?

E)What is the number of shares of the other stocks in this case?
utility
retail

Answer 1

Given,

	Computer (x)	Utility (y)	Retail (z)
Cost/share	30	44	26
Growth/share	6.00	6.00	2.40

a) By given condition we have, 30x+44y+26z = 392000...............(i)

6x+6y+(2.40)z = 58800

i.e., 5x+5y+2z = 49000................(ii)

Now, mutiplying (ii) by 6 and then subtracting from (i) we get, 14y+14z = 98000

i.e., y+z = 7000

i.e., y = 7000-z

Putting this value in (ii) we get, 5x+5(7000-z)+2z = 49000

i.e., 5x+35000-5z+2z = 49000

i.e., 5x-3z = 14000

i.e., x = 2800+(3/5)z

Therefore the solution is : Computer shares = 2800+(3/5)z, Utility shares = 7000-z, Retail shares = z.

b) Given, z = 1500.

Then, x = 2800+(3/5)*1500

i.e., x = 2800+900

i.e., x = 3700

And, y = 7000-1500

i.e., y = 5500

Therefore, Computer shares = 3700, Utility shares = 5500.

c) Given, z = 3000.

Then, x = 2800+(3/5)*3000

i.e., x = 2800+1800

i.e., x = 4600

And, y = 7000-3000

i.e., y = 4000

Therefore, Computer shares = 4600, Utility shares = 4000.

d) We have, x = 2800+(3/5)z

Which implies x is minimum when z is minimum ,i.e., z = 0. [Since x is an integer]

Then, x = 2800+(3/5)*0

i.e., x = 2800+0

i.e., x = 2800

And, y = 7000-0

i.e., y = 7000

Therefore, Computer shares = 2800, Utility shares = 7000, Retail shares = 0.

e) We have, x = 2800+(3/5)z and y = 7000-z.

Which implies that x is maximum when y is minimum.

Then, y is minimum means y = 0.

Now, 0 = 7000-z

i.e., z = 7000

And, x = 2800+(3/5)*7000

i.e., x = 2800+4200

i.e., x = 7000

Therefore, Computer shares = 7000, Utility shares = 0, Retail shares = 7000.