<PRE><B>Question 1120594</B>
A rectangular playground is to be fenced off and divided in two by another fence parallel to one side of the playground.
 452 feet of fencing is used.
 Find the dimensions of the playground that maximize the total enclosed area.
:
the dividing fence will be the width also; therefore
2L + 3W = 452
2L = -3W + 452
divide by 2
L = -1.5W + 226
:
Area = L * W
Replace L with (-1.5W+ 226)
A = W(-1.5W+226)
A = -1.5W^2 + 226W
A quadratic equation, max area occurs on the axis of symmetry, x= -b/(2a)
W = {{{(-226)/(2*-1.5)}}}
W = 75{{{1/3}}} ft is the Width
Find the Length
L = -1.5(75{{{1/3}}}) + 226
L = 113 ft is the length
Find Area
A = 113 * 75{{{1/3}}}
A = 8,512{{{2/3}}} sq ft


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