My Math Forum optimal number of shares

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 January 3rd, 2015, 11:23 AM #1 Newbie   Joined: Jan 2015 From: portland Posts: 1 Thanks: 0 optimal number of shares here is a problem: I have \$2000 I want to buy 2 different shares YYY (share price 109) and ZZZ (share price 14) what is the optimal number of YYY and ZZZ shares to buy to spend the most of \$2000 and have the remainder of that as close to zero as possible? 109y+14z=2000 what would be another equation in my Systems of equations? I cannot think of it. Please help. Last edited by greg1313; January 3rd, 2015 at 11:31 AM.
 January 3rd, 2015, 12:10 PM #2 Senior Member   Joined: Jan 2014 From: The backwoods of Northern Ontario Posts: 372 Thanks: 68 I am not sure that there is another equation. My thinking is that one should maximize the 14 dollar shares. To do that, I would buy just one 109 dollar share. This would leave 1891 to spend on the 14 dollar shares. 1891÷14=135.0717... If you buy 135 of these shares you will spend 1890 dollars. This amount when added to the 109 dollars sums to 1999 dollars. That's pretty darn close. But is there's an algebraic way to do it, I would like to become aware of it. And if there is another combination that comes closer to 2000 dollars, I would be surprised. Last edited by Timios; January 3rd, 2015 at 12:24 PM.
 January 3rd, 2015, 12:11 PM #3 Global Moderator   Joined: May 2007 Posts: 6,581 Thanks: 610 The only constraint you have is that x and y be non-negative integers and the left side cannot exceed the right side. By brute force try y between 0 and 18 to see what gets you the closest.
January 3rd, 2015, 05:08 PM   #4
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
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Hello, get42n8!

Quote:
 I have \$2000. I want to buy 2 different shares YYY (\$109 each) and ZZZ (\$14 each). What is the optimal number of YYY and ZZZ shares to buy to spend the most of \$2000 and have the remainder of that as close to zero as possible? $109y+14z\:=\:2000$

There is no second equation.
This is a Diophantine equation, requiring integer solutions.

This can be solved by Modular Arithmetic,
but here is an algebraic solution.

Solve for $z\!:\;\; z \;=\;\dfrac{2000-109y}{14} \quad\Rightarrow\quad z\;=\;142 - 8y + \dfrac{12+3y}{14}$

Since $z$ is an integer, $3(4+y)$ must be a multiple of 14.
The first time this happens is: $\:y = 10$

Hence: $\:z \:= \: \dfrac{2000-109(10)}{14} \:=\:\dfrac{910}{14} \:=\:65$

Therefore: $\:\boxed{y \,=\,10,\;z\,=\,65}$

Check: $\:109(10) + 14(65) \:=\:2000$

 January 3rd, 2015, 08:35 PM #5 Senior Member   Joined: Jan 2014 From: The backwoods of Northern Ontario Posts: 372 Thanks: 68 Good job, soroban! Please help me to understand why $z \;=\;\dfrac{2000-109y}{14} \quad\Rightarrow\quad z\;=\;142 - 8y + \dfrac{12+3y}{14}$
January 3rd, 2015, 11:50 PM   #6
Newbie

Joined: Jul 2012
From: Houston, Texas

Posts: 18
Thanks: 2

Math Focus: Geometry , Modern Algebra
Quote:
 Good job, soroban! Please help me to understand why $$z=\frac{2000−109y}{14}⇒z=142−8y+\frac{12+3 y}{14}$$
Because $$z = \frac{2000-109y}{14} = \frac{2000}{14} - \frac{109y}{14} \\ =142 + \frac{12}{14} - (8y - \frac{3y}{14}) \\ = 142 - 8y + \frac{12+3y}{14}$$

 January 5th, 2015, 11:21 AM #7 Newbie   Joined: Jan 2015 From: Turkie Posts: 9 Thanks: 2 This problem can be modeled as integer programming model as follows: Min x Subject to 109y+14z+x=2000 x,y,z >=0 and integer In operations research area, there are some methods which can be ussed to solve these models. Also some softwares like Lindo solver can also be used.

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