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

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 $$A$$?

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 $$A$$?

• Why would you not simplify the second equation by factoring out the $2$? – YiFan Jun 26 at 5:00
• The use of the recreational-mathematics tag intrigues me. Are you doing this problem for fun? Is it a smaller part of a larger problem? – Theo Bendit Jun 26 at 5:58
• It seems to me, that A can be as large as possible. – Ajay Mishra Jun 26 at 6:00
• They are just two non-parallel plane, they have to intersect. – Ajay Mishra Jun 26 at 6:01

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 $$a$$.

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 $$4A=49n-125$$.)