Question 844910
x for width and y for length;

The quantities of fence material will be 3 of y and 2 of x.  A picture would help to see this.  The sum of these quantities must be equal to the given amount of fence material, the 120 feet:  {{{2x+3y=120}}}.


The area {{{A=xy}}}.
Using the fence quantity equation, solve for y, and then substitute into the Area equation:
{{{3y=120-2x}}}
{{{y=40-(2/3)x}}};
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{{{A=x(40-(2/3)x)}}}
{{{highlight_green(A=40x-(2/3)x^2)}}}, a parabola with a maximum.
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The zeros can be found using the previous, factored form.  These zeros are at {{{x=0}}} and at {{{40-(2/3)x=0}}}.
{{{40=(2/3)x}}}
{{{x=40(3/2)}}}
{{{x=60}}}
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The maximum must occur in the exact middle of x=0 and x=60, which is at {{{highlight(x=30)}}}.
Maximum Area is {{{highlight(highlight(40*30-(2/3)*30^2))}}} square feet.