what is the minimum and maximum values of f ? f(x,y) = 2xy x^2 + 4y^2 = 32 So, t

Question

what is the minimum and maximum values of f ?

f(x,y) = 2xy

x^2 + 4y^2 = 32

So, the constraint g(x,y) = x^2 + 4y^2

fx = partial derivative of f with x
fx = 2y

fy = partial derivative f with y
fy = 2x

gx = 2x

gy = 8y

Now, using lagrange multiplier method,

fx = m * gx
2y = m*(2x)
y = mx

fy = m * gy
2x = m*(8y)
x = 4my

So, we have these conditions relationg x,y and m :

y = mx , x = 4my

x = 4m(mx)

x = 4m^2*x

4m^2 = 1

m^2 = 1/4

m = +/- 1/2

When m = 1/2 : y = x/2

x^2 + 4y^2 = 32

x^2 + 4(x/2)^2 = 32

x^2 + x^2 = 32

2x^2 = 32

x^2 = 16

x = +/- 4

When x = 4 , y = 4/2 --> y = 2
When x = -4 , y =-4/2 --> y = -2

Now, when m = -1/2 : y = mx ---> y = -x/2

x^2 + 4y^2 = 32
x^2 + 4(-x/2)^2 = 32
x^2 + 4(x^2/4) = 32
x^2 + x^2 = 32
2x^2 = 32
x^2 = 16
x = +/- 4

When x = 4 , y = -4/2 --> y = -2
When x = -4 , y = 4/2 --> y = 2

Explanation / Answer

we have got four critical points :

1. x = 4 , y = 4/2 --> y = 2 or (4, 2)

2. x = -4 , y =-4/2 --> y = -2 or (-4, -2)

3. x = 4 , y = -4/2 --> y = -2 or (4, -2)

4. x = -4 , y = 4/2 --> y = 2 or (-4, 2)

now, we plug all these four critival points in f(x,y) = 2xy

At (x,y)=(4,2)

f(4, 2)=2*4*2=16

At (x,y)=(-4,-2)

f(4, 2)=2*-4*-2=16

At (x,y)=(4,-2)

f(4, 2)=2*4*-2=-16

At (x,y)=(-4,2)

f(4, 2)=2*-4*2=-16

We can see that highest value among them is 16 and lowest value is -16

So, Maximum value of f is 16............Answer

And Minimum value of f is -16............Answer

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