Question 33204
Draw a straight line to denote the wall. Draw three more straight lines to create a rectangle.


Let 2 lengths of pen coming off the wall both be x.
The third length of te pen is therefore 36-2x in length.


Area, A = x(36-2x)


{{{ A = 36x-2x^2 }}}


Now there are 2 methods to find the maximum value.


1. Find the roots of this equation (where the curve crosses the x-axis). The maximum point on the curve will then lie equidistant from these two, since ALL parabolas are symmetric in shape.


2. Use differentiation to find the maximum value directly.


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1. {{{ 36x-2x^2 = 0}}}
{{{ x(36-2x) = 0}}}
so x=0 or 36-2x = 0
x=0 or 2x=36
x=0 or x=18


You can see this if i plot the graph: {{{graph(300,300,-4,22,-20,170,36x-2x^2)}}}


By knowing that a quadratic is symmetrical about the turning point, then roots are at x=0 and x=18, so the turning point, the maximum here, is at x=9.


So dimensions of the pen are 9x18. This makes the max area 162.


2.
{{{ A = 36x-2x^2 }}}
differentiate to find the turning point directly:
{{{ dA/dx = 36-4x }}}
{{{ 36-4x = 0}}}
4x = 36
--> x = 9


So, the maximum is at x=9. Area is as before


jon.