A RESPOSTAS QUESTIONÁRIO
Por: Elaine Chagas Abreu • 27/9/2020 • Trabalho acadêmico • 368 Palavras (2 Páginas) • 161 Visualizações
RESPOSTAS QUESTIONÁRIO
01 – A)
L(x)\ =\ -\frac{1}{8}\ x^{4\ \ }+\ \frac{17}{6}\ {x^3\ -\ 21x^2+\ 54x}^\
\bigmL(x)\prime\ =\ -\frac{x^3}{2}\ +\ \frac{17x^2}{2}\ {\ -\ 42\ x+\ 54}^\
L(x)\prime\ =\ \frac{{-\ x}^3\ +\ 17x^2}{2}\ \ {\ -\ 42\ x+\ 54}^\
x = 1
\frac{{-\ 1}^3\ +\ 17{.\ 1}^2}{2}\ \ {\ -\ 42\ .\ 1+\ 54}^\ =\ 8\ -\ 42\ +\ 54\ =\ 20
𝑥 = 2
\frac{{-2}^3\ +\ 17{.\ 2}^2}{2}\ \ {\ -\ 42\ .\ 2+\ 54}^\ =\ 30\ -\ 84\ +\ 54=\ 0
𝑥 = 4
\frac{{-4}^3\ +\ 17{.\ 4}^2}{2}\ \ {\ -\ 42\ .\ 4+\ 54}^\ =\ 104\ -168\ +\ 54=-10
𝑥 = 6
\frac{{-6}^3\ +\ 17{.\ 6}^2}{2}\ \ {\ -\ 42\ .\ 6+\ 54}^\ =\ 198\ -\ 252\ +\ 54\ =0
𝑥 = 8
\frac{{-8}^3\ +\ 17{.\ 8}^2}{2}\ \ {\ -\ 42\ .\ 8+\ 54}^\ =288\ -\ 336\ +54=\ 6
𝑥 = 9
\frac{{-9}^3\ +\ 17{.\ 9}^2}{2}\ \ {\ -\ 42\ .\ 9+\ 54}^\ =\ 324\ -\ 378\ +\ 54\ =\ 0
Ou seja, L(x)’ = 0 nos pontos 2, 6 e 9.
1 – B) \bigmL(x)\prime\prime\ =\ \frac{{-\ 3x}^2\ +\ 34x}{2}\ \ \ -\ 42
L(2)\prime\prime\ =\ \frac{{-\ 3\ .\ 2}^2\ +\ 34\ .\ 2}{2}\ \ - 42 = 28 - 42= -14 (máximo)
L(6)\prime\prime\ =\ \frac{{-\ 3\ .\ 6}^2\ +\ 34\ .\ 6}{2}\ \ - 42 =48 - 42= 6 (mínimo)
L(9)\prime\prime\ =\ \frac{{-\ 3\ .\ 9}^2\ +\ 34\ .\ 9}{2}\ \ {\ -\ 42\ \ =\ \ }^\ 31,5 – 42 = -10,5 (máximo)
01 – C)
L(x)\ =\ -\frac{1}{8}\ x^{4\ \ }+\ \frac{17}{6}\ {x^3\ -\ 21x^2+\ 54x}^\
L(2)\ =\ -\frac{1}{8}\ 2^{4\ \ }+\ \frac{17}{6}\ {2^3\ -\ 21{.\ 2}^2+\ 54\ .\ 2}^\ \ =-\ \frac{16}{8}+\frac{136}{6}\ -84\ +\ 108\ =\ 44,67
L(9)\ =\ -\frac{1}{8}\ 9^{4\ \ }+\ \frac{17}{6}\ 9^3\ -\ 21{.\ 9}^2+\ 54\ .\ 9=-\ \frac{6561}{8}+\frac{12393}{6}\ -1701+\ 486\ =\ 30,37
Ou seja, o lucro máximo foi na semana 2 de 44,67.
02 – A)
L\left(x\right)=42x-\left(-x^3\right)+22,5x^{2\ }-108x-100
L\left(x\right)\prime=42+\ 3x^2+45x^\ -108\ =\ \mathbf{3}\mathbit{x}^{\mathbf{2}\ \ }+\ \mathbf{45}\mathbit{x}\ -\ \mathbf{68}
\mathbf{3}\mathbit{x}^{\mathbf{2}\ \ }+\ \mathbf{45}\mathbit{x}\ -\ \mathbf{68}\ =\ \mathbf{0}
\frac{-b\pm\sqrt{b^2-4ac}}{2a}\ =\ \frac{-45\pm\sqrt{{45}^2-4.3.150}}{2.3}\ =\ \frac{-45\pm\sqrt{{45}^2-4.3.150}}{2.3}\ =\ \frac{-45\pm\sqrt{2025-1800}}{6}
\frac{-45\pm\sqrt{225}}{6}\ =\ \frac{-45\pm\ 15}{6}\ =\ \ \ x\prime=\frac{-45+\ 15}{6}\ =\ 5\ \ \ \ \ \ \ \ \ x\prime\prime=\ \frac{-45\ -\ 15}{6}\ =\ 10
Prova real: L(x) = \mathbf{3}\mathbit{x}^{\mathbf{2}\
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