Question 973099
Draw a rectangle, and make one dimension to be 20 units long, and the other dimension is made as unknown.


As given, 26 units  (or meters) of fencing.  AND  one of the sides of the garden is already taking 20 meters of the stone wall, so that will be 20 meters of fencing not needed for the garden.


Using x as the unknown dimenison of the garden,
account for the amount of fencing to use, {{{2x+20=26}}}.  
{{{x+10=13}}}
{{{x=3}}}



Intuitive sense tells you that you do not want to use a length for x less than 3, because this will only DECREASE the garden area; <s> you cannot increase x, because the calculation must made for for using all 20 meters of the stone wall.</s>
Say y and x are the dimensions.
{{{xy=A}}} for area;
{{{2x+y=26}}}, and {{{y<=20}}}, positive numbers only.
-


{{{y=26-2x}}}
substitute,
{{{x(26-2x)=A}}}
{{{highlight_green(A=-2x^2+26x)}}}


You want to find which x value is in the <b>exact middle of the roots</b> of the equation  {{{-2x^2+26x=0}}}, which will be the maximum value for where A is maximum.  


{{{-2x(x-13)=0}}}
Roots are 0 and 13.
Mid-value is at {{{x=6.5}}};
from this, {{{y=13}}}, for the side opposite from the stone wall.