1.5 X + 2.5 Y + 3 Z greater than or equal to T?

WithoutFlags

New member
Joined
Jan 19, 2018
Messages
2
Hello all,
I am new to the forum and I hope I am posting this question on right section.
If I am not, my deepest apologies.

So here goes:

We have:
a amount of stamps each valued 1 $, named x,
b amount of stamps each valued 1.4 $, named y,
c amount of stamps each valued 2.2 $, named z.

We are trying to reach a predefined sum of T with a combination of these stamps prefably getting an equal value, if we can't, a minimum higher value than T.
(Basically: ax + by + cz >= T)

How can we calculate this with minimum trial & error (none, if such thing is possible).
 
We have:
a amount of stamps each valued $1
b amount of stamps each valued $1.4
c amount of stamps each valued $2.2 $

We are trying to reach a predefined sum of T with a combination of these stamps preferably getting a value equal to T; if we can't, then we want the minimum value more than T.

Basically: 1a + 1.4b + 2.2c >= T

How can we calculate this with minimum trial & error (none, if such thing is possible).
You have one equation with four unknowns. It is not possible to "solve" this algebraically. All you can do is use guess-n-check. Give yourself some time.... ;)
 
We had a question very similar to this in the last question of the final exam for one of my math classes I took at ITT Technical institute. I told the instructor that I felt bad I couldn't figure it out and he basically told me I wasn't supposed to be able to figure it out.

I thought about this question a lot. What I tried to wrap my brain around is that to find the value of T in this case, you have to find where x, y, and z all meet at the same point. Ofc this may not even be a possibility because they could all be lines that never intersect, but you could try and graph all of these lines in 3 dimensional space and see if they do. Just as you have an x and y axis in two dimensional space, there is a z axis in 3 dimensional space.

But your answer is going to come out as three different terms on each axis (x, y, z), but that's only if there is even an answer to begin with.

If you take a point where there are two perpendicular walls and a ceiling that meets the walls, the corner of all three would be your intersecting (x, y, z) coordinates as a figure of example. Naturally, you are dealing with things that are not 3 dimensional, so you would just be using the idea of 3 dimensional space to solve for the amount of each stamp you would need and obviously round down to the nearest 1's place.

Or I am just crazy and I am not making any sense at all.
 
Last edited:
Hello all,
I am new to the forum and I hope I am posting this question on right section.
If I am not, my deepest apologies.

So here goes:

We have:
a amount of stamps each valued 1 $, named x,
b amount of stamps each valued 1.4 $, named y,
c amount of stamps each valued 2.2 $, named z.

We are trying to reach a predefined sum of T with a combination of these stamps prefably getting an equal value, if we can't, a minimum higher value than T.
(Basically: ax + by + cz >= T)

How can we calculate this with minimum trial & error (none, if such thing is possible).
Does the title have anything to do with the problem given in the question?

It is soluble without ANY trial and error because the minimization of the error provides the extra inequalities needed. We can deduce the algorithm below. I justify the first few steps in the algorithm.

\(\displaystyle \text {DEFINE } x = \lfloor t \rfloor \text { and } i = t - x.\)

NOTE: \(\displaystyle 0 \le i < 1.\)

\(\displaystyle x = 0 \implies a = 1,\ b = 0 = c.\)

Any other triplet of non-negative numbers will give a result greater than t by more than 0.99, which is the greatest excess possible with this triplet.

\(\displaystyle x > 0 \text { and } i = 0 \implies a = x,\ b = 0 = c.\) Obvious.

\(\displaystyle x = 1 \text { and } 0 < i \le 0.4 \implies b = 1,\ a = 0 = c.\)

Any other set will result in an excess over t greater 0.39, which is the greatest possible excess with this combination.

\(\displaystyle x = 1 \text { and } i > 0.4 \implies a = 2,\ b = 0 = c.\)

Anything else will result in an excess greater than 0.59.

\(\displaystyle x > 1 \text { and } 0 < i \le 0.2 \implies a = x - 2,\ b = 0,\ c = 2.\)

\(\displaystyle x > 1 \text { and } 0.2 < i \le 0.4 \implies a = x - 1,\ b = 1,\ c = 0.\)

\(\displaystyle x = 2 \text { and } 0.4 < i \le 0.6 \implies b = 2,\ a = 0 = c.\)

\(\displaystyle x > 2 \text { and } 0.4 < i \le 0.6 \implies a = x - 3,\ b = 1 = c.\)

\(\displaystyle x > 1 \text { and } 0.6 < i \le 0.8 \implies a = x - 2,\ b = 2,\ c = 0.\)

\(\displaystyle x > 1 \text { and } 0.8 < i \implies a = x + 1,\ b = 0 = c.\)

EDIT: Upon reflection, I am not sure that someone studying intermediate algebra should be expected to solve this problem. Ultimately the algorithm developed above can be justified algebraicaly, but I did not originally attack it that way. I attacked it by noticing that with the values given, I could get to within 19 cents of any given value of t once it was big enough. Then it was a matter of finding what was the meaning of big enough. This may be a good exercise in problem solving or discrete math (a field not very well defined in my opinion), but it is not easily approached from the standpoint of algebra.
 
Last edited:
Upon rereading my post from last night, I see that it is not as clear as it could be.

First of all, I assumed \(\displaystyle 100t \in \mathbb N^+.\)

Second, the student may not recognize the floor function.

\(\displaystyle 100t \in \mathbb N^+ \implies \exists\ x,\ j \text { such that } x,\ j \in \mathbb Z_{\ge 0}, xj > 0,\ j < 100, \text { and } t = x + \dfrac{j}{100}.\)

Given all that \(\displaystyle \lfloor t \rfloor = x \text { and } i = \dfrac{j}{100}.\)

Third, notice that at least one of a, b, and c must be positive and none can be negative.

Fourth, the algorithm itself varies with the value of t. It is perhaps clearest to start with the easy cases, when x > 2.

\(\displaystyle \text {I: }i = 0 \text { and } t = x + i \implies t = x + 0 = x * 1 \implies a = x,\ b = 0 = c.\)

\(\displaystyle \text {II: }0 < i \le 0.2 \text { and } t = x + i \implies x < t \le x + 0.2 \implies -\ x - 0.2 \le -\ t < -\ x \implies\)

\(\displaystyle 0 \le x + 0.2 - t < 0.2 \implies 0 \le \{(x - 2) * 1 + 2.2 * 1\} - t < 0.2 \implies\)

\(\displaystyle a = x - 2,\ b = 0,\ c = 1.\)

Using virtually identical logic, we get

\(\displaystyle \text{III: } 0.2 < i \le 0.4 \text { and } t = x + i \implies a = x - 1,\ b = 1,\ c = 0.\)

\(\displaystyle \text{IV: } 0.4 < i \le 0.6 \text { and } t = x + i \implies a = x - 3, b = 1,\ c = 2.\)

\(\displaystyle \text{V: } 0.6 < i \le 0.8 \text { and } t = x + i \implies a = x - 2,\ b = 2,\ c = 0.\)

\(\displaystyle \text{VI: } 0.8 < i < 1 \text { and } t = x + i \implies a = x + 1,\ b = 0 = c.\)

Now we look for exceptions. I will not work if x = 0 because then a = 0 so that a, b, and c will all be zero and thus less than positive t. But I will work if t is 1 or 2. So I gets split into 2 cases, x = 0 and x > 0.

II will not work if x < 2 because then a < 0, which is impossible. So II gets split into two cases, 0 < x < 2 and x > 1.

III will not work if x < 1 because then a < 0. So III gets split into two cases, 0 < x < 1 and x > 0.

IV will not work if x < 3. So it gets split, 0 < x < 3 and x > 2.

V will not work if x < 2. So it gets split, 0 < x < 2 and x > 1.
 
Last edited:
Top