Mathematics: He function f(x,y)equals5 x plus 5 y has an absolute maximum value and absolute minimum value subject to the constraint 9 x squared minus 9 xy plus 9 y squared equals 4. use lagrange multipliers to find these values.

Option (3)Step-by-step explanation: is a factor of a polynomial given in the options, that means a polynomial having factor as will be 0 for the value of x = .Option (1),3x⁴ + 26x² - 9= [For x = ]= 3(9i⁴) + 26(3i²) - 9 = 27 - 78 - 9 [Since i² = -1]= -60Option (2),4x⁴- 11x² + 3= = 4(9i⁴) - 33i² + 3= 36 + 33 + 3= 72Option (3),4x⁴ + 11x² - 3= = 4(9i⁴) + 33i² - 3= 36 - 33 - 3= 0Option (4),= = 3(9i⁴) - 26(3i²) - 9= 27 + 78 - 9= 96Therefore, is a factor of option (3)....

He function f(x,y)equals5 x plus 5 y has an absolute

Ответ:

MyaMya12

15.03.2022

Mathematics

You're looking for extrema of f(x,y)=5x+5y subject to 9x^2-9xy+9y^2=4 .

The Lagrangian is

$L(x,y,\lambda)=5x+5y+\lambda(9x^2-9xy+9y^2-4)$

with critical points where

$L_x=5+\lambda(18x-9y)=0\implies\dfrac5\lambda=9y-18x$

$L_y=5+\lambda(18y-9x)=0\implies\dfrac5\lambda=9x-18y$

$L_\lambda=9x^2-9xy+9y^2-4=0$

L_x=0 and L_y=0 together tell you that

$9y-18x=9x-18y\implies27x=27y\implies x=y$

Substituting y=x into $L_\lambda=0$ gives you

$9x^2-9x^2+9x^2-4=0\implies x^2=\dfrac49\implies x=\pm\dfrac23$

So you get two critical points, $\left(-\dfrac23,-\dfrac23\right)$ and $\left(\dfrac23,\dfrac23\right)$ .

$f\left(-\dfrac23,-\dfrac23\right)=-\dfrac{20}3$ (min)

$f\left(\dfrac23,\dfrac23\right)=\dfrac{20}3$ (max)

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Ответ:

waterborn9800

17.02.2021

Mathematics

Option (3)

Step-by-step explanation:

$(x-i\sqrt{3})$ is a factor of a polynomial given in the options, that means a polynomial having factor as $(x-i\sqrt{3})$ will be 0 for the value of x = $i\sqrt{3}$ .