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ollegr [7]
3 years ago
9

Consider the following equation. f(x, y) = e−(x − a)2 − (y − b)2 (a) Find the critical points. (x, y) = a,b (b) Find a and b suc

h that the critical point is at (−3, 8). a = b = (c) For the values of a and b in part (b), is (−3, 8) a local maximum, local minimum, or a saddle point?
Mathematics
1 answer:
murzikaleks [220]3 years ago
8 0

a.

f(x,y)=e^{-(x-a)^2-(y-b)^2}\implies\begin{cases}f_x=-2(x-a)e^{-(x-a)^2-(y-b)^2}\\f_y=-2(y-b)e^{-(x-a)^2-(y-b)^2}\end{cases}

Critical points occur where f_x=f_y=0. The exponential factor is always positive, so we have

\begin{cases}-2(x-a)=0\\-2(y-b)=0\end{cases}\implies(x,y)=\boxed{(a,b)}

b. As the previous answer established, the critical point occurs at (-3, 8) if \boxed{a=-3} and \boxed{b=8}.

c. Check the determinant of the Hessian matrix of f(x,y):

\mathbf H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}

The second-order derivatives are

f_{xx}=(-2+4(x-a)^2)e^{-(x-a)^2-(y-b)^2}

f_{xy}=4(x-a)(y-b)e^{-(x-a)^2-(y-b)^2}

f_{yx}=4(x-a)(y-b)e^{-(x-a)^2-(y-b)^2}

f_{yy}=(-2+4(y-b)^2)e^{-(x-a)^2-(y-b)^2}

so that the determinant of the Hessian is

\det\mathbf H(x,y)=f_{xx}f_{yy}-{f_{xy}}^2=\left((4(x-a)^2-2)(4(y-b)^2-2)-16(x-a)^2(y-b)^2\right)e^{-2(x-a)^2-2(y-b)^2}

\det\mathbf H(x,y)=(16(x-a)^2(y-b)^2-8(x-a)^2-8(y-b)^2)+4)e^{-2(x-a)^2-2(y-b)^2}

The sign of the determinant is unchanged by the exponential term so we can ignore it. For a=x=-3 and b=y=8, the remaining factor in the determinant has a value of 4, which is positive. At this point we also have

f_{xx}(-3,8;a=-3,b=8)=-2

which is negative, and this indicates that (-3, 8) is a local maximum.

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The speed of current will be "4.5 mph" and the rate Pratap can row in still water will be "8.5 mph".

What does "speed" mean in mathematics?

  • Speed is what it means. the speed of a change in an object's location in any direction.
  • Speed is defined as the ratio of distance to the amount of time it took to cover that distance.
  • Speed is a scalar quantity because it just has a direction and no magnitude.

Given:

Distance "26 miles" in time "2 hours".

Let,

Speed of water = y

Pratap speed when rowing in still water = x

As we know,

      Speed = distance/time

then,

  x + y = 26/2

  x + y = 13

 x = 13 - y

In return trip took him time "6.5 hours",

       x - y = 26/6.5

      x - y = 4

By substituting the value of "x", we get

 13 - y - y = 4

  13 - 2y = 4

     2y = 13 - 4

        2y = 9

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By substituting the value of "y", we get

    x = 13 - a

   x = 13 - 4. 5 = 8.5 mph (Pratap can row in still water)

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4 0
1 year ago
PLS HELPPP
Morgarella [4.7K]

Answer:

d. (9,-260°)

Step-by-step explanation:

100° = 100-360 = -260°

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Step-by-step explanation:

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