The graph of 2x + 3y = 8 is shown on the grid. Which ordered pair is in the solution set of 2x + 3y = 8?

Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If an answer d

andrey2020 [161]

Answer:

The maximum value is 1/27 and the minimum value is 0.

Step-by-step explanation:

Note that the given function is equal to $(xyz)^2$ then it means that it is positive i.e $f(x,y,z)\geq 0$ .

Consider the function $F(x,y,z,\lambda)=x^2y^2z^2-\lambda (x^2+y^2+z^2-1)$

We want that the gradient of this function is equal to zero. That is (the calculations in between are omitted)

$\frac{\partial F}{\partial x} = 2x(y^2z^2 - \lambda)=0$

$\frac{\partial F}{\partial y} = 2y(x^2z^2 - \lambda)=0$

$\frac{\partial F}{\partial z} = 2z(x^2y^2 - \lambda)=0$

$\frac{\partial F}{\partial \lambda} = (x^2+y^2+z^2-1)=0$

Note that the last equation is our restriction. The restriction guarantees us that at least one of the variables is non-zero. We've got 3 options, either 1, 2 or none of them are zero.

If any of them is zero, we have that the value of the original function is 0. We just need to check that there exists a value for lambda.

Suppose that x is zero. Then, from the second and third equation we have that

$-2y\lambda = -2z\lambda$ . If lambda is not zero, then y =z. But, since $-2y\lambda=0$ and lambda is not zero, this implies that x=y=z=0 which is not possible. This proofs that if one of the variables is 0, then lambda is zero. So, having one or two variables equal to zero are feasible solutions for the problem.

Suppose that only x is zero, then we have the solution set $y^2+z^2=1$ .

If both x,y are zero, then we have the solution set $z^2=1$ . We can find the different solution sets by choosing the variables that are set to zero.

NOw, suppose that none of the variables are zero.

From the first and second equation we have that

$\lambda = y^2z^2 = x^2z^2$ which implies $x^2=y^2$

Also, from the first and third equation we have that

$\lambda = y^2z^2 = x^2y^2$ which implies $x^2=z^2$

So, in this case, replacing this in the restriction we have $3z^2=1$ , which gives as another solution set. On this set, we have $x^2=y^2=z^2=\frac{1}{3}$ . Over this solution set, we have that the value of our function is $\frac{1}{3^3}= \frac{1}{27}$

4 0

3 years ago

In a trapezoid, the lower base is 15 cm, the upper base is 7 cm, and each of the two legs is 10 cm. How far must each of the leg

lora16 [44]

Answer:

15 feet apart

Step-by-step explanation:

brainliest?

4 0

3 years ago

How much money does he have

Andrej [43]

C $1.00

because 1/3 of 15 coins are 5 dimes

15 - 5 = 10 nickels

10 + 5 = 15 coins.

5 dimes= .50 cents

10 nickels= . 50 cents

.50 + .50 = $1.00

$1.00 is C.

*Hope this helped :)

7 0

3 years ago

Read 2 more answers

Michelle draws a card from a standard deck of 52 cards. She replaces the card and draws a second card. What is the probability t

erma4kov [3.2K]

To draw a heart, one would be choosing 1 card of 13 possible hearts, and 0 from the remaining 39 non-hearts. With respect to the entire deck, one would be choosing 1 card from 52 total cards. So the probability of drawing a heart is

$\dfrac{\dbinom{13}1\cdot\dbinom{39}0}{\dbinom{52}1}=\dfrac{13\cdot1}{52}=\dfrac14$

When Michelle replaces the card, the deck returns the normal, so the probability of drawing any card from a given suit is the same, $\dfrac14$ . In other words, drawing a spade is independent of having drawn the heart first.

So the probability of drawing a heart, replacing it, then drawing a spade is $\dfrac14\cdot\dfrac14=\dfrac1{16}$ .

4 0

3 years ago

Read 2 more answers

172.7=3.14d <br> Solve for d

Serga [27]

Answer:

55

Step-by-step explanation:

Divide both sides by 3.14 to get d by itself.

172.7 / 3.14 = 55 = d

4 0

3 years ago