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3 years ago

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Identify the variables, objective functions, and constraints

1 answer:

dsp733 years ago

7 0

What does it constraints?

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Use Lagrange multipliers to find the maximum and minimum values of (i) f(x,y)-81x^2+y^2 subject to the constraint 4x^2+y^2=9. (i

sp2606 [1]

i. The Lagrangian is

$L(x,y,\lambda)=81x^2+y^2+\lambda(4x^2+y^2-9)$

with critical points whenever

$L_x=162x+8\lambda x=0\implies2x(81+4\lambda)=0\implies x=0\text{ or }\lambda=-\dfrac{81}4$

$L_y=2y+2\lambda y=0\implies2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_\lambda=4x^2+y^2-9=0$

If $x=0$ , then $L_\lambda=0\implies y=\pm3$ .
If $y=0$ , then $L_\lambda=0\implies x=\pm\dfrac32$ .
Either value of $\lambda$ found above requires that either $x=0$ or $y=0$ , so we get the same critical points as in the previous two cases.

We have $f(0,-3)=9$ , $f(0,3)=9$ , $f\left(-\dfrac32,0\right)=\dfrac{729}4=182.25$ , and $f\left(\dfrac32,0\right)=\dfrac{729}4$ , so $f$ has a minimum value of 9 and a maximum value of 182.25.

ii. The Lagrangian is

$L(x,y,z,\lambda)=y^2-10z+\lambda(x^2+y^2+z^2-36)$

with critical points whenever

$L_x=2\lambda x=0\implies x=0$ (because we assume $\lambda\neq0$ )

$L_y=2y+2\lambda y=0\implies 2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_z=-10+2\lambda z=0\implies z=\dfrac5\lambda$

$L_\lambda=x^2+y^2+z^2-36=0$

If $x=y=0$ , then $L_\lambda=0\implies z=\pm6$ .
If $\lambda=-1$ , then $z=-5$ , and with $x=0$ we have $L_\lambda=0\implies y=\pm\sqrt{11}$ .

We have $f(0,0,-6)=60$ , $f(0,0,6)=-60$ , $f(0,-\sqrt{11},-5)=61$ , and $f(0,\sqrt{11},-5)=61$ . So $f$ has a maximum value of 61 and a minimum value of -60.

5 0

3 years ago

2,568 + (-1,452) pls help me asap !!! we are doing adding rational numbers

Kitty [74]

Answer:

1,116

Step-by-step explanation:

2568 + ( - 1452)

= 2568 - 1452 ( because + × - makes -)

= 2568 - 1452

=<u> 1,116</u>

4 0

3 years ago

Read 2 more answers

Use pascal's triangle to expand the following binomial expression <br>1.(2k-1/3)⁶

Gwar [14]

9514 1404 393

Answer:

64k^6 -64k^5 +(80/3)k^4 -(160/27)k^3 +(20/27)k^2 -(4/81)k +1/729

Step-by-step explanation:

The row of Pascal's triangle we need for a 6th power expansion is ...

1, 6, 15, 20, 15, 6, 1

These are the coefficients of the products (a^(n-k))(b^k) in the expansion of (a+b)^n as k ranges from 0 to n.

Your expansion is ...

1(2k)^6(-1/3)^0 +6(2k)^5(-1/3)^1 +15(2k)^4(-1/3)^2 +20(2k)^3(-1/3)^3 +...

15(2k)^2(-1/3)^4 +6(2k)^1(-1/3)^5 +1(2k)^0(-1/3)^6

= 64k^6 -64k^5 +(80/3)k^4 -(160/27)k^3 +(20/27)k^2 -(4/81)k +1/729

7 0

2 years ago

A) Simplify the expression and explain each step. (2 points)<br><br><br>4(3x+2) -2<br>= ?

TiliK225 [7]

Answer:

6 (2 x + 1)

Step-by-step explanation:

Simplify the following:

4 (3 x + 2) - 2

Hint: | Distribute 4 over 3 x + 2.

4 (3 x + 2) = 12 x + 8:

12 x + 8 - 2

Hint: | Group like terms in 8 + 12 x - 2.

Grouping like terms, 8 + 12 x - 2 = 12 x + (8 - 2):

12 x + (8 - 2)

Hint: | Subtract 2 from 8.

8 - 2 = 6:

12 x + 6

Hint: | Factor out the greatest common divisor of the coefficients of 12 x + 6.

Factor 6 out of 12 x + 6:

Answer: 6 (2 x + 1)

3 0

3 years ago

Prove that for a,b,c E ℤ, if a ❘ ( ab + c ), then a I c.

zhenek [66]

Answer:

True.

Step-by-step explanation:

Let a,b,c integers and a | (ab+c), that is to say that there exist a integer k such that ab+c = ka. Then:

ab+c = ka

c = ka-ab

c = a(k-b). So, c is a multiple of a, that is a | c.

3 0

3 years ago