Use the Method of Lagrange Multipliers to find the Minimum and Maximum of f(x, y) = xy subject to x^2+y^2-xy = 9 g

The Max fun is 9, and the Min fun is -3

Step-by-step explanation:

Please follow the steps carefully;

Let us consider the function f(x,y) = xy (1)

We will apply the Lagrange multipliers to maximize the function f(x,y) subject to the constraint  g(x,y) = x^2+y^2-xy = 9

By differentiating (1) w.r.t  x,   we get fx(x,y) = y

By differentiating (1) w.r.t  y,   we get fy(x,y) = x

By differentiating g(x,y) w.r.t  x,   we get gx(x,y) = 2x - y

Also By differentiating g(x,y) w.r.t  y,   we get gy(x,y) = 2y - x

let us take

fx = λgx

where y = λ(2x - y)

y/2x - y = λ   (2)

fy =  λgy

where x = λ(2y - x)

λ = x/2y -x (2)

Let us equate (2) and (3)

y/2x - y  = x/2y -x

2y² - xy = 2x² - xy

2y² = 2x² after cancelling like terms

y² = x²

So y = ±x

Now let us substitute y = x into the given constraint

x² + y² - xy = 9

x² + x² - x(x) = 9

x² = 9

therefore x = ± 3

We can conclude that when x = ±3, ⇒ y = ±3

The corresponding points are (3,3), (-3,-3)

Substitue y = -x in the given constraint gives

x² + y² - xy = 9

x² + (-x)² -x(-x) = 9

3x² = 9

x² = 3

x = ±√3

The corresponding points are (√3,-√3), (-√3,√3)

The function value is

f(x,y) = xy

f (√3,-√3) = (√3)(-√3) = -3

f (-√3,√3) = (-√3)(√3) = -3

we get;

f(3,3) = (3)(3) = 9

and

f (-3,-3) = (-3)(-3) = 9

We can conclusively say that ;

The Maximum value of the function is 9

The Minimum value of the function is -3

cheers i hope this helped