Question

alguém pode me ajudar por favor preciso que seja agora!

Answer 1

Questão 09.

$L(x,\,y)=60x+100y-\dfrac{3}{2}\,x^2-\dfrac{3}{2}\,y^2-xy$


A função $L$ é diferenciável em todos os pontos do plano. Logo, os candidatos a máximo/mínimo de $L$ são os pontos que anulam o vetor gradiente:

$\overrightarrow{\nabla}L(x,\,y)=\overrightarrow{0}\\\\ \left\langle\dfrac{\partial L}{\partial x}(x,\,y),\,\dfrac{\partial L}{\partial y}(x,\,y)\right\rangle=\left\langle0,\,0\right\rangle\\\\\\ \left\{ \!\begin{array}{l} \dfrac{\partial L}{\partial x}(x,\,y)=0\\\\ \dfrac{\partial L}{\partial y}(x,\,y)=0 \end{array} \right.$

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Calculando as derivadas parciais de $L:$

$\bullet\;\;\dfrac{\partial L}{\partial x}(x,\,y)=\dfrac{\partial}{\partial x}\!\left(60x+100y-\dfrac{3}{2}\,x^2-\dfrac{3}{2}\,y^2-xy \right )\\\\\\ \dfrac{\partial L}{\partial x}(x,\,y)=60-3x-y\\\\\\\\ \bullet\;\;\dfrac{\partial L}{\partial y}(x,\,y)=\dfrac{\partial}{\partial y}\!\left(60x+100y-\dfrac{3}{2}\,x^2-\dfrac{3}{2}\,y^2-xy \right )\\\\\\ \dfrac{\partial L}{\partial x}(x,\,y)=100-3y-x$

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Para que o vetor gradiente se anule, devemos ter

$\left\{\!\begin{array}{l} 60-3x-y=0\\\\ 100-3y-x=0 \end{array}\right.\\\\\\\\ \left\{\!\begin{array}{l} 3x+y=60\\\\ x+3y=100 \end{array}\right.$


Resolvendo o sistema acima, encontramos

$x=10~~\text{ e }~~y=30$


Logo, o único candidato a ponto de máximo/mínimo é o ponto $(10,\,30).$

Resposta: alternativa $\text{(C) }x=10~\text{ e }~y=30.$

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Questão 10.

$\displaystyle\int_R(x^2+y^2)\,dx\,dy$


sendo $R$ a região do plano descrita pelas seguintes condições:

$\left\{ \!\begin{array}{l} 0\le x\le 4\\\\ 0\le y\le \sqrt{x} \end{array} \right.~~~\Leftrightarrow~~~\left\{ \!\begin{array}{l} 0\le y\le 2\\\\ y^2\le x\le 4 \end{array} \right.$


Escrevendo as integrais iteradas (Teorema de Fubini), temos

$\displaystyle\int_R (x^2+y^2)\,dx\,dy=\int_0^2\int_{y^2}^4(x^2+y^2)\,dx\,dy\\\\\\ =\int_0^2 \left.\left(\dfrac{x^3}{3}+xy^2 \right )\right|_{x=y^2}^{x=4}\,dy\\\\\\ =\int_0^2\left[\left(\dfrac{4^3}{3}+4y^2 \right )-\left(\dfrac{(y^2)^3}{3}+(y^2)\cdot y^2 \right ) \right ]dy\\\\\\ =\int_0^2\left(\dfrac{64}{3}+4y^2-\dfrac{y^6}{3}-y^4\right )dy\\\\\\ =\left.\left(\dfrac{64y}{3}+\dfrac{4y^3}{3}-\dfrac{y^7}{21}-\dfrac{y^5}{5}\right )\right|_0^2$

$=\left(\dfrac{64\cdot 2}{3}+\dfrac{4\cdot 2^3}{3}-\dfrac{2^7}{21}-\dfrac{2^5}{5}\right )-0\\\\\\ =\dfrac{128}{3}+\dfrac{32}{3}-\dfrac{128}{21}-\dfrac{32}{5}\\\\\\ =\dfrac{4\,480}{105}+\dfrac{1\,120}{105}-\dfrac{640}{105}-\dfrac{672}{105}\\\\\\ =\dfrac{4\,480+1\,120-640-672}{105}\\\\\\ =\dfrac{4\,288}{105}$

$=\dfrac{4\,200+88}{105}\\\\\\ =\dfrac{4\,200}{105}+\dfrac{88}{105}\\\\\\ =\dfrac{105\cdot 40}{105}+\dfrac{88}{105}\\\\\\ =40+\dfrac{88}{105}\approx 41$


Resposta: alternativa $\text{(E) }41.$