Question

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.

Answer 1

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`