Can someone please help me on #2

Mathematics

1 answer:

Tom [10]3 years ago

4 0

1 - (-8)

Note that:

two positive = positive

one positive & one negative = negative

two negatives = positive

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Using the rules above, look at the question:

1 - (-8)

There are two negatives, so they change into positive. Add

1 - (-8) = 1 + 8 = 9

9 is your answer.

9 is the amount of spaces between -8 and 1

hope this helps

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I'm assuming the constraint involves some plus signs that aren't appearing for some reason, so that you're finding the extrema subject to $x^2+2y^2+3z^2=96$ .

Set $f(x,y,z)=xyz$ and $g(x,y,z)=x^2+2y^2+3z^2-96$ , so that the Lagrangian is

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$\begin{cases}\dfrac{xyz}2+\lambda x^2=0\\\\\dfrac{xyz}2+2\lambda y^2=0\\\\\dfrac{xyz}2+3\lambda z^2=0\\\\x^2+2y^2+3y^2=96\end{cases}$

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$\dfrac32xyz+96\lambda=0\implies \dfrac{xyz}2=-32\lambda$

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