<PRE><B>Question 1055297</B>
So then re-writing,
{{{y=(50-3x)/4}}}
{{{y=25/2-(3/4)x}}}
So then substituting,
{{{x+y=x+(25/2-(3/4)x)}}}
{{{x+y=25/2+x/4}}}
This function has a maximum value with the largest value of x that is still a positive integer.
{{{x=14}}} and {{{y=2}}} are the largest positive data pair and lead to the maximum sum.
{{{S[max]=14+2=16}}}
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*[illustration lp7.JPG].</PRE>