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Alla [95]
3 years ago
6

Find the point(s) on the surface z^2 = xy 1 which are closest to the point (7, 11, 0)

Mathematics
1 answer:
leonid [27]3 years ago
5 0
Let P=(x,y,z) be an arbitrary point on the surface. The distance between P and the given point (7,11,0) is given by the function

d(x,y,z)=\sqrt{(x-7)^2+(y-11)^2+z^2}

Note that f(x) and f(x)^2 attain their extrema, if they have any, at the same values of x. This allows us to consider the modified distance function,

d^*(x,y,z)=(x-7)^2+(y-11)^2+z^2

So now you're minimizing d^*(x,y,z) subject to the constraint z^2=xy. This is a perfect candidate for applying the method of Lagrange multipliers.

The Lagrangian in this case would be

\mathcal L(x,y,z,\lambda)=d^*(x,y,z)+\lambda(z^2-xy)

which has partial derivatives

\begin{cases}\dfrac{\mathrm d\mathcal L}{\mathrm dx}=2(x-7)-\lambda y\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dy}=2(y-11)-\lambda x\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dz}=2z+2\lambda z\\\\\dfrac{\mathrm d\mathcal L}{\mathrm d\lambda}=z^2-xy\end{cases}

Setting all four equation equal to 0, you find from the third equation that either z=0 or \lambda=-1. In the first case, you arrive at a possible critical point of (0,0,0). In the second, plugging \lambda=-1 into the first two equations gives

\begin{cases}2(x-7)+y=0\\2(y-11)+x=0\end{cases}\implies\begin{cases}2x+y=14\\x+2y=22\end{cases}\implies x=2,y=10

and plugging these into the last equation gives

z^2=20\implies z=\pm\sqrt{20}=\pm2\sqrt5

So you have three potential points to check: (0,0,0), (2,10,2\sqrt5), and (2,10,-2\sqrt5). Evaluating either distance function (I use d^*), you find that

d^*(0,0,0)=170
d^*(2,10,2\sqrt5)=46
d^*(2,10,-2\sqrt5)=46

So the two points on the surface z^2=xy closest to the point (7,11,0) are (2,10,\pm2\sqrt5).
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If Free Yer Assets Bank (FYAB) will give you 0.25% compounded quarterly and I.M.A.Q.T.π bank will give you 0.23% interest compou
ivann1987 [24]

Answer:

FYAB gives a better deal.

Step-by-step explanation:

Compound interest:

A = P(1 + \dfrac{r}{n})^{nt}

Continuously compounded interest:

A = Pe^{rt}

For the quarterly compounded interest, r = 0.25%, and n = 4.

2P = P(1 + \dfrac{0.0025}{4})^{4t}

1.000625^{4t} = 2

\log (1.000625^{4t}) = \log 2

4t(\log 1.000625) = \log 2

t = \dfrac{\log 2}{4\log 1.000625}

t = 277

For the continuously compounded interest, r =0.23%

A = Pe^{rt}

2P = Pe^{0.0023t}

2 = e^{0.0023t}

\ln 2 = \ln e^{0.0023t}

\ln 2 = 0.0023t

t = \dfrac {\ln 2}{0.0023}

t = 301

The quarterly compounded doubles in 277 years.

The continuously compounded doubles in 301 years.

Answer: FYAB gives a better deal.

7 0
3 years ago
1. Nikola invested $6.000 in a 3 year
Kamila [148]
I would rather say A or C
3 0
2 years ago
What is the 30th term of the linear sequence below?<br><br> -5,-7,-9,-11,-13....
adelina 88 [10]

Answer:

-63

Step-by-step explanation:

This arithmetic progression

The formula is a+(n-1)d

a means the first value

n is the nth term

d is the difference

So a is -5

n is 30

d is -2

So let's substitute

-5+(30-1)-2

-5+(29)-2

-5-58

-63

Therefore the final answer is-63

Just follow the step and the general formula, you will get your final answer

3 0
3 years ago
I need help with this please
Tanya [424]

Angle 1 and 2 are <u>supplementary</u> angles.

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3 years ago
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Last year, Maria had $10,000 to invest. She invested some of it in an account that paid 6% simple interest per year, and she inv
denis-greek [22]

Answer:

$3,000 invested at 6%

$7,000 invested at 10%

Step-by-step explanation:

Maria had total $10,000 to invest.

Let x be the amount that Maria invested initially at 6% interest rate.

0.06x

Then she invested the remaining amount at 10% interest rate.

0.10(10,000 - x)

She received a total of $880 in interest.

0.06x + 0.10(10,000 - x) = 880

0.06x + 1000 - 0.10x = 880

-0.04x = 880 - 1000

-0.04x = -120

0.04x = 120

x = 120/0.04

x = $3,000

This is the amount that Maria invested initially at 6% interest rate.

The remaining amount is

10,000 - 3,000

$7,000

This is the remaining amount that she invested at 10% interest rate.

3 0
3 years ago
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