Question 133489
So far, so good.


{{{A=L*(15-L)}}} is absolutely correct.  But you probably want to put this into standard form:


{{{A=-L^2+15L}}}


This is just a quadratic in the form {{{f(x)=ax^2+bx+c}}}, substituting L for the independent variable x and with coefficients as follows:


{{{a=-1}}}
{{{b=15}}}
{{{c=0}}}


The vertex of the parabola has an x-coordinate at {{{-b/2a}}}, so {{{-15/2(-1)=15/2}}}


The y-coordinate of the vertex is equal to the value of the function at the x-coordinate, so {{{-(15/4)^2+15(15/2)=-225/4+225/2=225/4}}}


Hence the vertex of the graph is at (15/2,225/4).


Factors of {{{-L^2+15L}}} are {{{L(15-L)}}}, so zeroes of the polynomial are 0 and 15, meaning the x-intercepts are at (0,0) and (15,0).


Pick a couple other values for L:


If L = 1, then {{{A=-(1)^2+15(1)=14}}}, and by symmetry, if L = 14, A = 14


If L = 2, then {{{A=-(2)^2+15(2)=26}}}, and by symmetry, if L = 13, A = 26


Just plot the points and sketch a smooth curve.


{{{drawing(600,600,-5,20,-5,60,
grid(1),
graph(600,600,-5,20,-5,60,-x^2+15x),
circle(15/2,225/4,.2),
circle(0,0,.2),
circle(15,0,.2),
circle(1,14,.2),
circle(14,14,.2),
circle(2,26,.2),
circle(13,26,.2)
)}}}


Super-Double-Plus Extra Credit.  Just looking at this picture, what specific quadrilateral gives the maximum area for a given perimeter?