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Natali [406]
4 years ago
5

Please help me on this

Mathematics
1 answer:
Lena [83]4 years ago
4 0
y+x=8\\
y=8-x\\\\
\dfrac{12}{x}=8-x\qquad x\not=0\\
12=8x-x^2\\
x^2-8x+12=0\\
x^2-8x+16-4=0\\
(x-4)^2=4\\
x-4=2 \vee x-4=-2\\
x=6 \vee x=2\\\\
y=8-6 \vee y=8-2\\
y=2 \vee y=6\\
\boxed{(x,y)=\{(2,6),(6,2)\}}


\displaystyle
S_P=\int\limits_2^68-x-\dfrac{12}{x}\, dx\\
S_P=\left[8x-\dfrac{x^2}{2}-12\ln x\right]_2^6\\
S_P=8\cdot6-\dfrac{6^2}{2}-12\ln 6-(8\cdot2-\dfrac{2^2}{2}-12\ln 2)\\
S_P=48-18-12\ln 6-16+2+12\ln2\\
S_P=16-12\ln 6+12\ln 2\\
S_P=16-12(\ln 6-\ln 2)\\\boxed{S_P=16-12\ln 3}

\displaystyle
S_Q=\int\limits_2^6\dfrac{12}{x}\, dx\\
S_Q=\left[12\ln x\right]_2^6\\
S_Q=12\ln 6-12\ln 2\\
S_Q=12(\ln6-\ln 2)\\
\boxed{S_Q=12\ln 3}

\displaystyle
V=\pi \int \limits_a^b f^2(x)\, dx\\\\
V=\pi \int \limits_2^6 \left(\dfrac{12}{x}\right)^2\, dx\\
V=\pi \int \limits_2^6 \dfrac{144}{x^2}\, dx\\
V=\pi\left[-\dfrac{144}{x}\right]_2^6\\
V=\pi \cdot \left(-\dfrac{144}{6}-\left(-\dfrac{144}{2}\right)\right)\\
V=\pi \cdot (-24+72)\\
V=\pi \cdot48\\
\boxed{V=48\pi}
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