Find the maximum and minimum values of the function f(x,y)=4x^2 +9y^2
subject to xy = 4. Use Lagrange multipliers.
This are I did
{1. fx=8x = gx=λy, 2. fy=18y =gx=λx, 3. xy=4
(8x/18y)=(λy/λx)------->8x^2 =18y^2-------->8x^2 -18y^2=0. I don't know what to do next.
So far, so good. Now include the constraint xy=4
4x^2 = 9y^2
4x^2 = 9(4/x)^2
x^2 = 36/x^2
x^4 = 36
x = ±√6
y = ±2/3 √6
So the minimum f(x,y) = 48
This may seem strange, since the minimum value of 4x^2+9y^2 is clearly 0. But with the constraint xy=4, we see that our collection of ellipses is tangent to the ellipse xy=4 only for the ellipse 4x^2+9y^2 = 48.
A nice discussion of the topic is presented here:
https://tutorial.math.lamar.edu/classes/calciii/lagrangemultipliers.aspx
To find the maximum and minimum values of the function f(x,y) = 4x^2 + 9y^2 subject to the constraint xy = 4 using Lagrange multipliers, we need to consider the Lagrangian function L(x, y, λ) defined as:
L(x, y, λ) = f(x, y) - λ(g(x, y)), where g(x, y) is the constraint equation.
In this case, our function f(x, y) is 4x^2 + 9y^2, and the constraint equation is xy = 4.
Let's begin by calculating the partial derivatives of f(x, y) with respect to x and y:
∂f/∂x = 8x
∂f/∂y = 18y
Now, let's calculate the partial derivatives of the constraint equation g(x, y) = xy = 4 with respect to x and y:
∂g/∂x = y
∂g/∂y = x
Using these partial derivatives, we can set up the system of equations to solve for the critical points:
∂L/∂x = 8x - λy = 0 (Equation 1)
∂L/∂y = 18y - λx = 0 (Equation 2)
g(x, y) = xy - 4 = 0 (Equation 3)
To solve this system, we eliminate λ by multiplying Equation 1 by x and Equation 2 by y:
(8x^2 - λxy) = 0 (Equation 4)
(18y^2 - λxy) = 0 (Equation 5)
Now, we can add Equation 4 and Equation 5:
8x^2 + 18y^2 - λxy - λxy = 0
Simplifying, we get:
8x^2 + 18y^2 - 2λxy = 0 (Equation 6)
We also use Equation 3 (g(x, y) = xy - 4 = 0), which gives:
xy - 4 = 0 (Equation 7)
Now, let's solve this system of equations (Equation 6 and Equation 7) to find the critical points:
From Equation 7, we can isolate x or y by solving for x:
x = 4/y
Substituting this into Equation 6, we get:
8(4/y)^2 + 18y^2 - 2λ(4/y)y = 0
Simplifying, we have:
32/y^2 + 18y^2 - 8λ = 0
Multiplying through by y^2 to eliminate the denominator, we get:
32 + 18y^4 - 8λy^2 = 0 (Equation 8)
Now, multiplying Equation 7 by y, we get:
xy^2 - 4y = 0
Substituting x = 4/y, we have:
4y^2 - 4y = 0
4y(y - 1) = 0
This gives two possible solutions:
1) y = 0
2) y - 1 = 0, which gives y = 1
Substituting these values back into Equation 8, we can solve for λ:
For y = 0:
32 + 18(0)^4 - 8λ(0)^2 = 0
32 - 8λ = 0
λ = 4
For y = 1:
32 + 18(1)^4 - 8λ(1)^2 = 0
32 + 18 - 8λ = 0
50 - 8λ = 0
λ = 50/8
λ = 25/4
Now that we have the values of λ for the critical points, we can substitute them back into Equations 1 and 2 to solve for x:
For y = 0 and λ = 4:
8x - 4(0) = 0
8x = 0
x = 0
For y = 1 and λ = 25/4:
8x - (25/4)(1) = 0
8x - 25/4 = 0
8x = 25/4
x = 25/32
Finally, we can substitute the values of x and y into the original function f(x, y) = 4x^2 + 9y^2 to find the maximum and minimum values:
For x = 0 and y = 0:
f(0, 0) = 4(0)^2 + 9(0)^2
f(0, 0) = 0
For x = 25/32 and y = 1:
f(25/32, 1) = 4(25/32)^2 + 9(1)^2
f(25/32, 1) = 625/256 + 9
f(25/32, 1) = 625/256 + 2304/256
f(25/32, 1) = 2929/256
Therefore, the minimum value of the function f(x, y) = 4x^2 + 9y^2 subject to xy = 4 is 0, and the maximum value is 2929/256.
To find the maximum and minimum values of the function f(x, y) = 4x^2 + 9y^2 subject to the constraint xy = 4, we can use the method of Lagrange multipliers. Here's how you can do it step by step:
1. Define the Lagrangian function L(x, y, λ) as the sum of the original function f(x, y) and the constraint multiplied by the Lagrange multiplier λ:
L(x, y, λ) = f(x, y) - λ(xy - 4)
2. Calculate the partial derivatives of L(x, y, λ) with respect to x, y, and λ:
∂L/∂x = 8x - λy
∂L/∂y = 18y - λx
∂L/∂λ = -(xy - 4)
3. Set the partial derivatives equal to zero and solve the resulting system of equations:
8x - λy = 0 ...(1)
18y - λx = 0 ...(2)
xy - 4 = 0 ...(3)
4. Solve equations (1) and (2) simultaneously to eliminate the Lagrange multiplier λ. Multiply equation (1) by 18 and equation (2) by 8 to get:
144x - 18λy = 0 ...(4)
144y - 8λx = 0 ...(5)
Multiply equation (5) by 18 and equation (4) by 8 to get:
1152y - 144λx = 0 ...(6)
1152x - 144λy = 0 ...(7)
Add equations (6) and (7) to eliminate λ:
1152x + 1152y = 0
Divide by 1152:
x + y = 0 ...(8)
5. Substitute equation (8) into equation (3) to get the values of x and y:
(x)(-x) - 4 = 0
x^2 = -4
x = ±2i, where i is the imaginary unit
Since we are considering real solutions, there are no real values for x that satisfy equation (3).
6. To determine the maximum and minimum values, we need to consider the endpoints of the constraint. Since there are no real solutions for x, we can only consider the case where y approaches infinity or negative infinity.
As y approaches infinity, f(x, y) = 4x^2 + 9y^2 will approach infinity.
As y approaches negative infinity, f(x, y) = 4x^2 + 9y^2 will also approach infinity.
Therefore, there are no maximum or minimum values of f(x, y) subject to the given constraint.
In summary, the function f(x, y) = 4x^2 + 9y^2 subject to the constraint xy = 4 does not have maximum or minimum values.