Find the maximum value of $a-b$ where $4a+5b=61$ and where $a$ and $b$ are positive whole numbers.

Question

I am at a beginners level at maths. I have tried to work out the question myself, but I have seen a more simpler way of getting to the answer in which I haven't understood.

Here's the simpler approach to this question:

$4a+5b=61$

When $b=1$, $a=14$

When $b=5$, $a=9$

When $b=9$, $a=4$

From the pattern we add $4$ to the value given to b each time and we subtract $5$ from the value given to a each time. That is the shortcut to get all the possible values.

Why is this so?

The solution goes on to say when $a=14$ and $b=1$ ; $a-b=13$ which is the maximum value of $a-b$.

Answer 1

We have the diophantine equation

$$4a+5b=61$$

To see that every integer solution can be produced by taking a particular solution and then adding some multiple of $5$ to $a$ and subtracting the same multiple of $4$ from $b$, consider the following:

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$$4a=61-5b$$

$$5b=61-4a$$

We see (where '$|$' means 'divides)

$$4\,|\,61-5b$$

$$5\,|\,61-4a$$

$\implies$

$$4\,|\,1-b$$

$$5\,|\,1+a$$

$\implies$

$$4\,|\,b-1$$

$$5\,|\,a+1$$

Hence if $(a_0, b_0)$ is a solution, any other solution must be of the form $(a_0+5i, b_0+4j)$ for some integers $i, j$.

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Next, we need to show if $(a_0, b_0)$ is a solution, then $(a_0+5k, b_0-4k)$ is also a solution for all integers $k$.

$$4a_0+5b_0=61$$

$\implies$

$$4a_0+5b_0+20k-20k=61$$

$\implies$

$$4a_0+5b_0+4(5k)-5(4k)=61$$

$$4(a_0+5k)+5(b_0-4k)=61$$

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Therefore, all solutions must be of the form $(a, b)=(a_0+5k, b_0-5k)$ once a particular solution $(a,b)=(a_0,b_0)$ is found and for all integers $k$.

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However, the question is after positive integer solutions, hence the largest value of $a-b$ is when $(a,b)=(14,1)$

Answer 2

HINT: $$b =\frac{61}{5}-\frac{4a}{5}$$ then $$a-b=\frac{9}{5}a-\frac{61}{5}$$

Answer 3

Since $61(-4+5)=61$, we can extend this to the full solution set with $$ 4\ \overbrace{(5k-61)}^a+5\ \overbrace{(61-4k)}^b=61 $$ For $a,b\ge0$, we need $13\le k\le15$. Since $a-b=9k-122$, we want to use $k=15$ where $a-b=13$.

Answer 4

$$\max_{\begin{array}{c} 4a+5b=61 \\ a,b\in \mathbb Z^+ \end{array}}a-b$$

A solution to $4a+5b=61$ is $a=9$ and $b=5$. So the general solution is $a=9-5t$ and $b=5+4t$, where $t$ is an integer.

We note that $b> 0 \implies t \ge -1$ and $a > 0 \implies t \le 1$. So we need $-1 \le t \le 1$.

Since $a-b = 4-9t$ is a decreasing function of $t$, its maximum value will occur when $t=-1$. That is when $(a,b) = (14, 1)$