Question 87522
Hi,<br />

If last year, the width of the garden was {{{x}}}, then there would be {{{x^2}}} trees. If the garden is {{{a}}} units wider this year, then there are {{{(x+a)^2}}} trees this year. The difference between this year and last year is 11 trees. So {{{(x+a)^2-x^2=11}}}. If we expand that:<br />

{{{2xa+a^2=11}}}

Now, we had some trees last year so we know {{{x>0}}} and we have more trees this year so {{{a>0}}}. Also the number of trees is integer, so both {{{x}}} and {{{a}}} must be integers (this is easy to prove).<br />

Now looking at the equation, we can see that {{{a<4}}} this is because a value of 4 or greater would force {{{x}}} to be negative so a can only be 1, 2, or 3.<br />

If a is 1, x is 5, if a is 2, x is 7/4, and if a is 3, x is 1/3. The only integer solution here is (1,5). Which gives a crop size (number of trees) this year of (1+5)*(1+5)=36<br />

Hope that helps,
Kev