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aivan3 [116]
3 years ago
12

What is the distance between (-1,4) and (2,0)

Mathematics
1 answer:
Ivahew [28]3 years ago
7 0

Answer: 5

The Distance between the coordinates of two points is measured by distance formula.

Step-by-step explanation:

The Distance between the coordinates of two points P(x1,y1) and Q(x2,y2) is measured by

Distance PQ= \sqrt{}(x2-x1)²+(y2-y1)²

So, the distance between (-1,4 ) and (2,0) can be calculated by

Distance= \sqrt{}(2-(-1))²+(0-4)²

              = \sqrt{}(3)²+(-4)²

              =\sqrt{}9+16 = \sqrt{}25                                                  as    \sqrt{}25 = 5

Distance = 5

So the distance between the two coordinates is 5.

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Find the minimum and maximum of f(x,y,z)=x2+y2+z2 subject to two constraints, x+2y+z=7 and x−y=6.
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Use the method of Lagrange multipliers. We have Lagrangian

L(x,y,z,\lambda_1,\lambda_2)=x^2+y^2+z^2+\lambda_1(x+2y+z-7)+\lambda_2(x-y-6)

with partial derivatives (set equal to 0) of

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-7=0
L_{\lambda_2}=x-y-6=0

As x+2y+z=7, and x-y=6, we can obtain

\dfrac12L_x+L_y+\dfrac12L_z=0\implies3\lambda_1-\dfrac12\lambda_2=-7
L_x-L_y=0\implies\lambda_1-2\lambda_2=12
\begin{cases}3\lambda_1-\frac12\lambda_2=-7\\\lambda_1-2\lambda_2=12\end{cases}\implies\lambda_1=-\dfrac{40}{11},\lambda_2=-\dfrac{86}{11}

From this, we find a single critical point:

2x-\dfrac{40}{11}-\dfrac{86}{11}=0\implies x=\dfrac{63}{11}
\dfrac{63}{11}-y=6\implies y=-\dfrac3{11}
\dfrac{63}{11}-\dfrac6{11}+z=7\implies z=\dfrac{20}{11}

At this point, we have a value of

f\left(\dfrac{63}{11},-\dfrac3{11},\dfrac{20}{11}\right)=\dfrac{398}{11}

To determine what kind of extremum occurs at this point, we check the Hessian of f(x,y,z):

\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}2&0&0\\0&2&0\\0&0&2\end{bmatrix}

We observe that \det\mathbf H(x,y,z)=8>0 at any point (x,y,z), and that the eigenvalues of this matrix are all positive (2 with multiplicity 3), so \mathbf H is positive definite. By the second partial derivative test, this means f(x,y,z) attains a minimum at this critical point. Meanwhile, f has no maximum value.
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3 years ago
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