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The function ​f(x,y)equals=3 x plus 3 y 3 x+3y has an absolute maximum value and absolute minimum value subject to the constraint 9 x squared minus 9 xy plus 9 y squared equals 169 x 2−9xy+9y2=16. Use Lagrange multipliers to find these values.

User Jberryman
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1 Answer

4 votes

Answer:

Therefore the absolute maximum value is
f(\frac43,\frac43)=8

the absolute minimum value is
f(-\frac43,-\frac43)=-8

Explanation:

Method of Lagrange multipliers:

1. Solve the following system of equation


\bigtriangledown L(x,y,..,\lambda) =\bigtriangledown f(x,y,..)+\lambda \bigtriangledown g(x,y,...)=0

2. Plug the all solution (x,y,..) from the first step into f(x,y,..) and identify the maximum and minimum values.

The constant
\lambda is known as the Lagrange Multiplier.

Given function is

f(x,y)= 3x+3y

subject to constrain
9x^2-9xy+9y^2=16


g(x)=9x^2-9xy+9y^2-16=0

The partial derivatives are


(\partial f)/(\partial x)=3


(\partial f)/(\partial y)=3


(\partial g)/(\partial x)=18x-9y


(\partial g)/(\partial y)=-9x+18y


\therefore (\partial L)/(\partial x)=3+\lambda (18x-9y)=0......(1)


\therefore (\partial L)/(\partial y)=3+\lambda (-9x+18y)=0......(2)


\therefore (\partial L)/(\partial \lambda)=9x^2-9xy+9y^2-16=0.....(3)

Multiplying 2 with equation (2) and add with equation (1)


3+ 18\lambda x-9\lambda y +6-18\lambda x+36\lambda y=0


\Rightarrow 9+27 \lambda y=0


\Rightarrow y=-(9)/(27 \lambda)


\Rightarrow y=-(1)/(3 \lambda)

Putting the value of y in equation (1)


3+ 18\lambda x-9\lambda (-(1)/(3\lambda))=0


\Rightarrow 3+18\lambda x+3=0


\Rightarrow x=-(6)/(18\lambda)


\Rightarrow x=-(1)/(3 \lambda)

Putting the value of x and y in equation (3)


9(-(1)/(3\lambda))^2-9(-(1)/(3\lambda))(-(1)/(3\lambda))+9(-(1)/(3\lambda))^2-16=0


\Rightarrow 9((1)/(9\lambda^2))-9((1)/(9\lambda^2))+9((1)/(9\lambda^2))-16=0


\Rightarrow 1-1+1-16\lambda^2=0


\Rightarrow \lambda^2=(1)/(16)


\Rightarrow \lambda=\pm(1)/(4)

Therefore the value of x and y becomes


x=\pm(1)/(3((1)/(4)))


\Rightarrow x=\pm (4)/(3)

and
y=\pm (4)/(3)

Therefore


f((4)/(3),\frac43) =(3* \frac43)+(3* \frac43)

=4+4

=8


f(-(4)/(3),-\frac43) =[3* (-\frac43)]+[3*(- \frac43)]

=-4-4

=-8

Therefore the absolute maximum value is
f(\frac43,\frac43)=8

the absolute minimum value is
f(-\frac43,-\frac43)=-8

User Will Pierce
by
6.5k points
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