The coordinates of quadrilateral ABCD plotted on a coordinate plane are A(0, 3), B(3,-1).

Mathematics

1 answer:

Vikki [24]2 years ago

3 0

Answer:

24.56 (parallelogram)

Step-by-step explanation:

4 sides of ABCD: AB, BC, AD, DC

perimeter = AB + BC + AD + DC

distance of two points: √(x-x')²+(y-y')²

AB = √3²+4² = √25 = 5

BC = √7²+2² = √53 = 7.28

AD = √7²+2² = √53 = 7.28

DC = √3²+4² = √25 = 5

P = 5*2 + 7.28*2 = 24.56

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iogann1982 [59]

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The Hessian matrix for this function looks like

$H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}-18e^y&-18xe^y\\-18xe^y&9e^y(2-x^2+4y+y^2)\end{bmatrix}$

and has determinant

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Maxima occur whenever the determinant is positive and $f_{xx}$ . Minima occur whenever both the determinant and $f_{xx}$ are positive. Saddle points occur whenever the determinant is negative.

At $(0,0)$ , you have a saddle point since the determinant reduces to -324, so $(0,0,0)$ is the saddle point.

At $(0,-2)$ , the determinant is $\dfrac{324}{e^4}>0$ and $f_{xx}(0,-2)=-\dfrac{18}{e^2}$ , so $\left(0,-2,\dfrac{36}{e^2}\right)$ is a local maximum.

No other critical points remain, so you're done.

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2 years ago

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sashaice [31]

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Step-by-step explanation: