Please help!!!!!!solve

Mathematics

1 answer:

fgiga [73]2 years ago

5 0

I’m a bit confused confused, but the answer may be -2x^3 + 2x^2 - 2x +2

Work:

To solve this, take the equation for q and the equation for r, and multiply them together.

Use the distributive property to multiply them.

I’m not sure if this is right or not, so pls don’t get mad if it isn’t.

Thank you! :)

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Find the minimum and maximum of f(x,y,z)=x^2+y^2+z^2 subject to two constraints, x+2y+z=4 and x-y=8.

Alika [10]

The Lagrangian for this function and the given constraints is

$L(x,y,z,\lambda_1,\lambda_2)=x^2+y^2+z^2+\lambda_1(x+2y+z-4)+\lambda_2(x-y-8)$

which has partial derivatives (set equal to 0) satisfying

$\begin{cases}L_x=2x+\lambda_1+\lambda_2=0\\L_y=2y+2\lambda_1-\lambda_2=0\\L_z=2z+\lambda_1=0\\L_{\lambda_1}=x+2y+z-4=0\\L_{\lambda_2}=x-y-8=0\end{cases}$

This is a fairly standard linear system. Solving yields Lagrange multipliers of $\lambda_1=-\dfrac{32}{11}$ and $\lambda_2=-\dfrac{104}{11}$ , and at the same time we find only one critical point at $(x,y,z)=\left(\dfrac{68}{11},-\dfrac{20}{11},\dfrac{16}{11}\right)$ .

Check the Hessian for $f(x,y,z)$ , given by

$\mathbf H(x,y,z)=\begin{bmatrix}f_{xx}&f_{xy}&f_{xz}\\f_{yx}&f_{yy}&f_{yz}\\f_{zx}&f_{zy}&f_{zz}\end{bmatrix}=\begin{bmatrix}=\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}$

$\mathbf H$ is positive definite, since $\mathbf v^\top\mathbf{Hv}>0$ for any vector $\mathbf v=\begin{bmatrix}x&y&z\end{bmatrix}^\top$ , which means $f(x,y,z)=x^2+y^2+z^2$ attains a minimum value of $\dfrac{480}{11}$ at $\left(\dfrac{68}{11},-\dfrac{20}{11},\dfrac{16}{11}\right)$ . There is no maximum over the given constraints.

7 0

3 years ago

In the xy-plane, the equations x + 2y = 10 and

konstantin123 [22]

By solving given equations, the value of c is 30.

Given two equations

x + 2y = 10 and

3x + 6y = c

These lines represents the same line for some constant c.

Value of c:

x + 2y = 10-------------(1)

3x + 6y = c-------------(2)

Dividing equation (2) by 3

$\frac{3x}{3}+ \frac{6y}{3} =\frac{c}{3}$

After solving the above equation, we get

x + 2y = c/3-----------(3)

Remember that a line is written as ax + by = c, in our case, both lines have a =1 and b = 2. Therefore, in orther that the two lines are equal, we need that, 10 = c/3

c = 10 × 3 = 30

c = 30

Therefore,

The value of c is 30.

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7 0

1 year ago

What two represents quantities that are proportional A.30/36 and 40/46. B.40/48 and 100/120. C.4/9 and 14/19. D. 27/81 and 21/72

lubasha [3.4K]

The answer is 4/9 and 14/19