$ 4x -5y + 24z = 4A $, $ 2x - 2y + 2z = 10$. What is the largest possible value of $A$?

Question

If $x,y,z$ integers that satisfy $$ 4x -5y + 24z = 4A $$ $$ 2x - 2y + 2z = 10$$ with $y < 2x$ and $y-20z< 0$, what is the largest possible value of $A$?

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Attempt:

We can rewrite the equations as:

$$ -y + 20z = 4A - 20 $$ $$ 2x - 2y + 2z = 10$$

or $$ -y + 20z = 4A - 20$$ $$ x - 19z = 25 - 4A $$

so we get $y - 20z = 20 - 4A < 0 $, so $A > 5$. Also, we can obtain

$$ -y + 20z = 4A - 20 $$ $$ 2x - (19/10)y = 12 - (2/5)A$$ so we have general solutions $$ z = A/5 -1 + y/20$$ $$ x = 6 - A/5 + (19y/20) $$ how to continue find the maximum possible value of $A$?

Answer 1

Solving the system $$ \begin{cases} 4x-5y+24z=a \qquad\qquad\;\;\; \\[4pt] 2x-2y+2z=10\\ \end{cases} $$ for $x,y$ yields $$ \begin{cases} x=19z+25-4a \qquad\qquad\;\;\; \\[4pt] y=20z+20-4a\\ \end{cases} \\[18pt] $$ \begin{align*} \text{Then}\;\;&y - 20z < 0\\[4pt] \iff\;&(20z+20-4a)-(20z) < 0\\[4pt] \iff\;&20-4a < 0\\[4pt] \iff\;&a > 5\\[4pt] \iff\;&a \ge 6\\[10pt] \text{and}\;\;&y < 2x\\[4pt] \iff\;&20z+20-4a < 2(19z+25-4a)\\[4pt] \iff\;&10z+10-2a < 19z+25-4a\\[4pt] \iff\;&9z+15 > 2a\\[4pt] \iff\;&z > \frac{2a-15}{9}\\[4pt] \end{align*} Thus, for any integer $a \ge 6$, and any integer $z > {\large{\frac{2a-15}{9}}}$, we can let $$ \begin{cases} x=19z+25-4a \qquad\qquad\;\;\; \\[4pt] y=20z+20-4a\\ \end{cases} $$ and all the specified conditions are satisfied.

Hence, while there is a minimum value for $a$, namely $a=6$, there is no maximum value for $a$.

Answer 2

Eliminate $y$ with $y=x+z-5$. The constraints become

$$-x+z<5,\\x-19z<5$$ and the objective function is

$$4A=4x-5y+25z=29x+20y-125.$$

As the constraints include an unbounded part of the first quadrant, $A$ is unbounded above !

(For instance, $x=n,y=2n-5,z=n$ satisfy the constraints and yield $4A=49n-125$.)