Lote\u00a0B<\/strong><\/td>\n| 60<\/td>\n | 10<\/td>\n | 40<\/td>\n | 50<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n As inc\u00f3gnitas<\/strong> do problema s\u00e3o:<\/p>\n\n- x<\/em>, o n\u00famero de lotes A adquiridos<\/li>\n
- y<\/em>, o n\u00famero de lotes B adquiridos<\/li>\n<\/ul>\n
Organizemos os dados numa tabela para podermos relacion\u00e1-los mais facilmente.<\/p>\n \n\n\nLote de flores<\/strong><\/td>\nN.\u00ba de lotes comprados<\/strong><\/td>\nN.\u00ba de flores adquiridas (unidades)<\/strong><\/td>\nDespesa (\u20ac)<\/strong><\/td>\n<\/tr>\n\nJacintos<\/strong><\/td>\nT\u00falipas\u00a0<\/strong><\/td>\nNarcisos<\/strong><\/td>\n<\/tr>\n\nLote A<\/strong><\/td>\n| $x$<\/td>\n | $30x$<\/td>\n | $40x$<\/td>\n | $30x$<\/td>\n | $75x$<\/td>\n<\/tr>\n | \nLote B<\/strong><\/td>\n| $y$<\/td>\n | $10y$<\/td>\n | $40y$<\/td>\n | $50y$<\/td>\n | $60y$<\/td>\n<\/tr>\n | \nTotal<\/strong><\/td>\n| $x+y$<\/td>\n | $30x+10y$<\/td>\n | $40x+40y$<\/td>\n | $30x+50y$<\/td>\n | $75x+60y$<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n \n O problema tem as seguintes restri\u00e7\u00f5es<\/strong>:<\/p>\n\n- \n
$x\\in {{\\mathbb{N}}_{0}}$ e $y\\in {{\\mathbb{N}}_{0}}$ \nO n\u00famero de lotes a adquirir de cada tipo \u00e9 inteiro n\u00e3o negativo;<\/div>\n<\/li>\n - \n
$30x+10y\\ge 1200\\Leftrightarrow 3x+y\\ge 120$ \nS\u00e3o necess\u00e1rios, pelo menos, 1200 jacintos;<\/div>\n<\/li>\n - $40x+40y\\ge 3200\\Leftrightarrow x+y\\ge 80$
\nS\u00e3o necess\u00e1rias, pelo menos, 3200 t\u00falipas;<\/li>\n - $30x+50y\\ge 3000\\Leftrightarrow 3x+5y\\ge 300$
\nS\u00e3o necess\u00e1rios, pelo menos, 3000 narcisos.<\/li>\n<\/ol>\nCada lote A \u00e9 vendido a 75 \u20ac e cada lote B \u00e9 vendido a 60 \u20ac.
\nPortanto, a fun\u00e7\u00e3o objetivo<\/strong> \u00e9: $D=75x+60y\\Leftrightarrow y=-\\frac{5}{4}x+\\frac{D}{60}$<\/p>\nOs pontos admiss\u00edveis, isto \u00e9, os pontos cujas coordenadas satisfazem as condi\u00e7\u00f5es impostas, s\u00e3o os pontos de coordenadas inteiras pertencentes ao dom\u00ednio plano representado, no seu interior ou sobre a sua fronteira.<\/p>\n Validemos as coordenadas dos pontos G e H, assinalados no referencial apresentado:<\/p>\n $\\begin{matrix}
\n\\left\\{ \\begin{array}{*{35}{l}}
\n3x+y=120\u00a0 \\\\
\nx+y=80\u00a0 \\\\
\n\\end{array} \\right. & \\Leftrightarrow\u00a0 & \\left\\{ \\begin{array}{*{35}{l}}
\n2x=40\u00a0 \\\\
\nx+y=80\u00a0 \\\\
\n\\end{array} \\right. & \\Leftrightarrow\u00a0 & \\left\\{ \\begin{array}{*{35}{l}}
\nx=20\u00a0 \\\\
\ny=60\u00a0 \\\\
\n\\end{array} \\right. & \\to\u00a0 & H(20,60)\u00a0 \\\\
\n\\end{matrix}$<\/p>\n $\\begin{matrix}
\n\\left\\{ \\begin{array}{*{35}{l}}
\nx+y=80\u00a0 \\\\
\n3x+5y=300\u00a0 \\\\
\n\\end{array} \\right. & \\Leftrightarrow\u00a0 & \\left\\{ \\begin{array}{*{35}{l}}
\n2y=60\u00a0 \\\\
\nx+y=80\u00a0 \\\\
\n\\end{array} \\right. & \\Leftrightarrow\u00a0 & \\left\\{ \\begin{array}{*{35}{l}}
\nx=50\u00a0 \\\\
\ny=30\u00a0 \\\\
\n\\end{array} \\right. & \\to\u00a0 & G(50,30)\u00a0 \\\\
\n\\end{matrix}$<\/p>\n<\/p>\n | | | | | | | | | | | |