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4 years ago

13

The formula for the indicated variable

1 answer:

Gala2k [10]4 years ago

3 0

Can't answer, I have no idea

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The equations that must be solved for maximum or minimum values of a differentiable function w=f(x,y,z) subject to two constrai

sammy [17]

The Lagrangian is

$L(x,y,z,\lambda,\mu)=x^2+y^2+z^2+\lambda(4x^2+4y^2-z^2)+\mu(2x+4z-2)$

with partial derivatives (set equal to 0)

$L_x=2x+8\lambda x+2\mu=0\implies x(1+4\lambda)+\mu=0$

$L_y=2y+8\lambda y=0\implies y(1+4\lambda)=0$

$L_z=2z-2\lambda z+4\mu=0\implies z(1-\lambda)+2\mu=0$

$L_\lambda=4x^2+4y^2-z^2=0$

$L_\mu=2x+4z-2=0\implies x+2z=1$

Case 1: If $y=0$ , then

$4x^2-z^2=0\implies4x^2=z^2\implies2|x|=|z|$

Then

$x+2z=1\implies x=1-2z\implies2|1-2z|=|z|\implies z=\dfrac25\text{ or }z=\dfrac23$

$\implies x=\dfrac15\text{ or }x=-\dfrac13$

So we have two critical points, $\left(\dfrac15,0,\dfrac25\right)$ and $\left(-\dfrac13,0,\dfrac23\right)$

Case 2: If $\lambda=-\dfrac14$ , then in the first equation we get

$x(1+4\lambda)+\mu=\mu=0$

and from the third equation,

$z(1-\lambda)+2\mu=\dfrac54z=0\implies z=0$

Then

$x+2z=1\implies x=1$

$4x^2+4y^2-z^2=0\implies1+y^2=0$

but there are no real solutions for $y$ , so this case yields no additional critical points.

So at the two critical points we've found, we get extreme values of

$f\left(\dfrac15,0,\dfrac25\right)=\dfrac15$ (min)

and

$f\left(-\dfrac13,0,\dfrac23\right)=\dfrac59$ (max)

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4 years ago

3. What is the perimeter of Patrick's<br> figure?

Rina8888 [55]

Answer : I need a graph or something

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4 years ago

EXPERTS/ACE/GENIUSES

shusha [124]

3 is the scale factor you would use to dilate Figure A to Figure B.

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3 years ago

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<br><br><br><br> <br><br> , find f^{-1}(x)f <br><br>−1<br><br> (x)

Stolb23 [73]

Answer:

Multiply the numerator by the reciprocal of the denominator. Multiply x x by 1 1 . Since g(f(x))=x g ( f ( x ) ) = x , f−1(x)=1x f - 1 ( x ) = 1 x is the inverse of f(x)=1x f ( x ) = 1 x .

Step-by-step explanation:

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2 years ago

A total of 428 tickets were sold for the school play. They were either adult tickets or student tickets. The number of student t

Liula [17]

425 tickets, if only 3 student tickets were sold... 428 total minus 3 student tickets = 425 adult tickets + 3 sold student tickets.

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3 years ago