Question

Urgente: AB/CD/EFGH = 07/04/1990
Questão: As funções de receita marginal e custo marginal de uma empresa são Rmg (q) = 3q2 – (AC + BD)q + AGB e Cmg (q) = -3q2 + (GAB + CD)q + FCH, onde a variável "q" representa a quantidade e as funções Receita e o Custo são representadas em unidades monetárias. Determine o que se pede:
a) (0,4 pontos) A função custo, sabendo que para q=AB temos custo igual a EBF unidades monetárias;

b) (0,4 pontos) A variação total do custo no intervalo A ≤ q ≤ED

c) (0,6 pontos) Os pontos de máximo local, mínimo local, e ponto de inflexão, da Função Custo, se existirem;

d) (0,4 pontos) A função receita;

e) (0,4 pontos) A função lucro;

f) (0,4 pontos) O intervalo onde o lucro é crescente;

g) (0,4 pontos) O lucro mínimo local, se existir.

Answer 1

A partir da data de nascimento, vamos especificar os valores para as letras:

$A = 0\ \ \ \ \ \ B=7\ \ \ \ \ C=0\ \ \ \ \ D=4\\ E=1\ \ \ \ \ \ F=9\ \ \ \ \ G=9\ \ \ \ \ H=0\\\\\\ Fun\c{c}\~ao\ Receita\ marginal:\\\\ R_{mg(q)}=3q^2-(07+74)q+097\\\\ \boxed{R_{mg(q)}=3q^2-81q+97}\\\\\\ Fun\c{c}\~ao\ Custo\ marginal:\\\\ C_{mg(q)} = -3q^2+(907+04)q+900\\\\ \boxed{C_{mg(q)} = -3q^2+911q+900}$



$a)\ Para\ q=07\ temos\ custo\ igual\ a\ 179\\\\ Integrando\ fun\c{c}\~ao\ custo\ marginal\ para\ encontrar\\\ a\ fun\c{c}\~ao\ custo:\\\\ C_{(q)}=\displaystyle \int -3q^2+911q+900\\\\ C_{(q)}=-3\frac{q^{2+1}}{2+1}+911\frac{q^{1+1}}{1+1}+900\frac{q^{0+1}}{0+1}+constante\\\\ C_{(q)}=-q^3+911\frac{q^{2}}{2}+900q+constante\\\\\\ Encontrando\ o\ valor\ da\ constante:\\\\ 179=-(7)^3+911\frac{(7)^2}{2}+900(7)+constante\\\\ 179=-343+\frac{44.639}{2}+6.300+constante$


$constante = 179+343-\frac{44.639}{2}-6.300\\\\ constante = -28.097,5\\\\\\ Fun\c{c}\~ao\ custo\ completa:\\\\ \boxed{C_{(q)}=-q^3+911\frac{q^{2}}{2}+900q-28.097,5}$



$b)\ Intervalo\ 0 \leq q \leq 14\\\\ Para\ q=0:\\\\ C_{(0)}=-(0)^3+911\frac{(0)^{2}}{2}+900(0)-28.097,5\\\\ C_{(0)}=R\ \ 28.097,50\\\\\\ Para\ q=14:\\\\ C_{(14)}=-(14)^3+911\frac{(14)^{2}}{2}+900(14)-28.097,5\\\\ C_{(14)}=-2.744+911(98)+12.600-28.097,5\\\\ C_{(14)}=-2.744+89.278+12.600-28.097,5\\\\ C_{(14)}=R\ \ 71.036,50\\\\\\ Intervalo:\\\\ \boxed{28.097,50 \leq q \leq 71.036,50}$



$c)\ Encontrando\ as\ raizes\ para\ C_{mg(q)}=-3q^2+911q+900\\\\ Termos:\ \ \ \ a=-3\ \ \ b=911\ \ \ c=900\\\\ Usando\ bhaskara:\\\\ q=\dfrac{-b\pm \sqrt{b^2-4.a.c}}{2.a}\\\\ q=\dfrac{-911\pm \sqrt{(911)^2-4.(-3).(900)}}{2.(-3)}\\\\ q=\dfrac{911\pm \sqrt{829.921+10.800}}{6}\\\\ q=\dfrac{911\pm \sqrt{840.721}}{6}\\\\ q\approx q=\dfrac{911\pm 916,91}{6}\\\\ q\approx 151,83\pm 152,82\\\\ \boxed{q'\approx304,65\ \ \ \ \ \ \ q''\approx-0,99}$


$Encontrando\ o\ custo\ minimo:\\\\ Para\ q=-0,99:\\\\ C_{(-0,99)}=-(-0,99)^3+911\frac{(-0,99)^{2}}{2}+900(-0,99)-28.097,5\\\\ C_{(-0,99)}\approx0,97+911\frac{0,98}{2}-891-28.097,5\\\\ \boxed{C_{(-0,99)}\approx-28.541,14}\\\\\\ Encontrando\ o\ custo\ maximo:\\\\ Para\ q=304,65:\\\\ C_{(304,65)}=-(304,65)^3+911\frac{(304,65)^{2}}{2}+900(304,65)-28.097,5\\\\ C_{(304,65)}\approx-28.275.060,79+42.275.694,05+274.185-28.097,5\\\\ \boxed{C_{(304,65)}=R\ \ 14.246.720,76}$


$Pontos\ de\ Inflex\~ao:\\\\ P_1=(-0,99\ \ \ ,\ \ \ -28.541,14)\\\\ P_2=(304,65\ \ \ ,\ \ \ 14.246.720,76)$



$d)\ Integrando\ fun\c{c}\~ao\ Receita\ marginal\ para\ encontrar\\ a\ fun\c{c}\~ao\ receita:\\\\ R_{(q)} = \displaystyle \int 3q^2-81q+97\\\\ R_{(q)} = 3\frac{q^{2+1}}{2+1}-81\frac{q^{1+1}}{1+1}+97\frac{q^{0+1}}{0+1}+constante\\\\ R_{(q)} = q^3-81\frac{q^2}{2}+97q+constante$

$OBS:\ Constante\ ser\'a\ igual\ a\ zero,\ porque\ sem\ quantidade\\ vendida,\ n\~ao\ h\'a\ receita\\\\\\ Fun\c{c}\~ao\ Receita\ completa:\\\\ \boxed{R_{(q)} = q^3-81\frac{q^2}{2}+97q}$



$e)\ Fun\c{c}\~ao\ Lucro:\\\\ L_{(q)} = R_{(q)} - C_{(q)}\\\\ L_{(q)} = (q^3-81\frac{q^2}{2}+97q) - (-q^3+911\frac{q^{2}}{2}+900q-28.097,5)\\\\ L_{(q)} = q^3-81\frac{q^2}{2}+97q +q^3-911\frac{q^{2}}{2}-900q+28.097,5\\\\ \boxed{L_{(q)} = 2q^3-992\frac{q^2}{2}-803q+28.097,5}$



$f)\ Intervalo\ do\ lucro\ crescente:\\\\ L_{mg(q)}=R_{mg(q)} - C_{mg(q)}\\\\ L_{mg(q)}=(3q^2-81q+97) - (-3q^2+911q+900)\\\\ L_{mg(q)}=3q^2-81q+97 +3q^2-911q-900\\\\ L_{mg(q)}=6q^2-992q-803\\\\\\ Termos:\ \ \ \ a=6\ \ \ \ b=-992\ \ \ \ c=-803\\\\ Encontrando\ as\ raizes\ usando\ Bhaskara:\\\\ q=\dfrac{-b \pm \sqrt{b^2-4.a.c}}{2.a}\\\\ q=\dfrac{-(-992) \pm \sqrt{(-992)^2-4.(6).(-803)}}{2.(6)}\\\\ q=\dfrac{992 \pm \sqrt{984.064+19.272}}{12}\\\\ q\approx\dfrac{992 \pm 1.001,67}{12}$


$q\approx82,67 \pm 83,47\\\\ q'=-0,8\ \ \ \ \ q''=166,14\\\\\\ Intervalo\ crescente\ do\ lucro:\\\\ \boxed{S=\{q \in \mathbb{R}\ \ //\ \ (-\infty \ \textless \ q \leq-0,8 )\ \ e\ \ (166,14 \leq q \ \textless \ \infty ) \}}\\\\\\ PS:\ Pode\ ser\ comprovado\ no\ gr\'afico\ em\ anexo.$



$g)\ N\~ao\ existe\ lucro\ minimo\ local$


Espero ter ajudado.
Bons estudos!