Question 1068827
Let {{{ W }}} = width ( 3 equal sizes )
Let {{{ L }}} = length ( 2 equal sizes )
{{{ 200 = 3W + 2L }}}
{{{ 2L = 200 - 3W }}}
{{{ L = 100 - (3/2)*W }}}
---------------------------
(a)
Let {{{ A }}} = the total area
{{{ A = W*L }}}
{{{ A = W*( 100 - (3/2)*W ) }}}
{{{ A = -(3/2)*W^2 + 100W }}}
----------------------------
(b)
This is a parabola with a maximum due
to the minus sign. 
The W-value of the function maximum is:
{{{ W[max] = -b/(2a) }}}
{{{ W[max] = (-100)/(2*(-3/2) ) }}}
{{{ W[max] = 100/3 }}}
and
{{{ L[max] = 100 - (3/2)*W }}}
{{{ L[max] = 100 - (3/2)*(100/3) }}} 
{{{ L[max] = 100 - 50 }}}
{{{ L[max] = 50 }}}
----------------------------
 The lengths that give the maximum area are:
{{{ W = 33.333
{{{ L = 50 }}}
-----------------
check:
{{{ 200 = 3W + 2L }}}
{{{ 200 = 3*33.333 + 2*50 }}}
{{{ 200 = 100 + 100 }}}
and
{{{ A = W*L }}}
{{{ A = 33.333*50 }}}
{{{ A = 1666.667 }}}
( note that you can replace {{{ W }}} with {{{ x }}} and
{{{ A }}} with {{{ y }}}, and that will give {{{ y = f(x) }}}
the way they want. )
Here's the plot of {{{ A = f(W) }}}
{{{ graph( 400, 400, -10, 100, -200, 2000, -(3/2)*x^2 + 100x ) }}}