![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/12pag223-75.jpg\" "\\"Par\u00e1bola\\"")
<\/a><\/p>\nA \u00e1rea do ret\u00e2ngulo [EFGH] \u00e9 dada, com $0 < x < 3$, por: $$\\begin{array}{*{20}{l}}
\n{A(x)}& = &{2 \\times \\overline {OH}\u00a0 \\times \\overline {HE} } \\\\
\n{}& = &{2 \\times x \\times \\left( { – {x^2} + 9} \\right)} \\\\
\n{}& = &{2x\\left( {9 – {x^2}} \\right)}
\n\\end{array}$$<\/p>\n
Ora, $$\\begin{array}{*{20}{l}}
\n{A'(x)}& = &{\\left( {2x\\left( {9 – {x^2}} \\right)} \\right)’} \\\\
\n{}& = &{\\left( {18x – 2{x^3}} \\right)’} \\\\
\n{}& = &{18 – 6{x^2}}
\n\\end{array}$$<\/p>\n
Como $18 – 6{x^2} = 0 \\Leftrightarrow {x^2} = 3 \\Leftrightarrow x =\u00a0 \\pm \\sqrt 3 $, temos:<\/p>\n