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3 years ago

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Help me please Im the big not smart

1 answer:

Tems11 [23]3 years ago

8 0

Answer:

Step-by-step explanation:

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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)

5 0

3 years ago

Write a decimal that represents the value of one 1.00 bill and 5 quarters

oee [108]

$1.00

5 quarters=$1.25

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3 years ago

I need help and I’m marking as brainliest and if u do thank you

liubo4ka [24]

Answer:

1:2

Step-by-step explanation:

25:50 can be simplified down to 1:2. Think about ratios like fractions, since 25 can be divided into 50 two times, 25 is 1/2 of 50.

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IrinaK [193]

Answer:

6-16y

Step-by-step explanation:

2(3-8y)

3(2)-8y(2)

6-16y

please mark brainliest and have a great day!

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elena-14-01-66 [18.8K]

.07 or seven hundredths

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