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Natasha_Volkova [10]
2 years ago
13

Let the (x; y) coordinates represent locations on the ground. The height h of

Mathematics
1 answer:
grigory [225]2 years ago
7 0

The critical points of <em>h(x,y)</em> occur wherever its partial derivatives h_x and h_y vanish simultaneously. We have

h_x = 8-4y-8x = 0 \implies y=2-2x \\\\ h_y = 10-4x-12y^2 = 0 \implies 2x+6y^2=5

Substitute <em>y</em> in the second equation and solve for <em>x</em>, then for <em>y</em> :

2x+6(2-2x)^2=5 \\\\ 24x^2-46x+19=0 \\\\ \implies x=\dfrac{23\pm\sqrt{73}}{24}\text{ and }y=\dfrac{1\mp\sqrt{73}}{12}

This is to say there are two critical points,

(x,y)=\left(\dfrac{23+\sqrt{73}}{24},\dfrac{1-\sqrt{73}}{12}\right)\text{ and }(x,y)=\left(\dfrac{23-\sqrt{73}}{24},\dfrac{1+\sqrt{73}}{12}\right)

To classify these critical points, we carry out the second partial derivative test. <em>h(x,y)</em> has Hessian

H(x,y) = \begin{bmatrix}h_{xx}&h_{xy}\\h_{yx}&h_{yy}\end{bmatrix} = \begin{bmatrix}-8&-4\\-4&-24y\end{bmatrix}

whose determinant is 192y-16. Now,

• if the Hessian determinant is negative at a given critical point, then you have a saddle point

• if both the determinant and h_{xx} are positive at the point, then it's a local minimum

• if the determinant is positive and h_{xx} is negative, then it's a local maximum

• otherwise the test fails

We have

\det\left(H\left(\dfrac{23+\sqrt{73}}{24},\dfrac{1-\sqrt{73}}{12}\right)\right) = -16\sqrt{73} < 0

while

\det\left(H\left(\dfrac{23-\sqrt{73}}{24},\dfrac{1+\sqrt{73}}{12}\right)\right) = 16\sqrt{73}>0 \\\\ \text{ and } \\\\ h_{xx}\left(\dfrac{23+\sqrt{73}}{24},\dfrac{1-\sqrt{73}}{12}\right)=-8 < 0

So, we end up with

h\left(\dfrac{23+\sqrt{73}}{24},\dfrac{1-\sqrt{73}}{12}\right)=-\dfrac{4247+37\sqrt{73}}{72} \text{ (saddle point)}\\\\\text{ and }\\\\h\left(\dfrac{23-\sqrt{73}}{24},\dfrac{1+\sqrt{73}}{12}\right)=-\dfrac{4247-37\sqrt{73}}{72} \text{ (local max)}

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<u>Step-by-step explanation</u>:

ABCD is a quadrilateral with their opposite sides are congruent (equal).

The both pairs of opposite sides are given as AB = 3 + x , DC = 4x , AD = y + 1 , BC = 2y.

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Keep y terms on one side and constant on other side.

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If the inicial value is 4, we have:

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3: f(1) = -1/4(1) + 1 = 3/4

4: f(3/4) = -1/4(3/4) + 1 = 13/16

5: f(13/16) = -1/4(13/16) + 1 = 51/64

6: f(51/64) = -1/4(51/64) + 1 = 205/256

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6: f(411/512) = -1/4(411/512) + 1 = 1637/2048

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8: f(6555/8192) = -1/4(6555/8192) + 1 =  26213/32768

9: f(26213/32768) = -1/4(26213/32768) + 1 = 104859/131072

10: f(104859/131072) = -1/4(104859/131072) + 1 = 419429/524288 = 0.8

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