Find the maximum and minimum values of f(x,y)=3x+y on the ellipse x^2+4y^2=1
To find the maximum and minimum values of the function f(x, y) = 3x + y on the ellipse x^2 + 4y^2 = 1, we can use the method of Lagrange multipliers.
First, let's set up the problem using Lagrange multipliers. We need to find critical points where the gradient of the function, ∇f(x, y), is parallel to the gradient of the constraint function, ∇g(x, y), multiplied by a scalar λ.
The gradient of f(x, y) is:
∇f(x, y) = (3, 1)
The constraint function is:
g(x, y) = x^2 + 4y^2 - 1
Taking the gradient of the constraint function:
∇g(x, y) = (2x, 8y)
Now we set up the equations:
∇f(x, y) = λ * ∇g(x, y)
(3, 1) = λ * (2x, 8y)
This gives us two equations:
3 = 2λx
1 = 8λy
We also have the equation of the ellipse constraint: x^2 + 4y^2 = 1
Now, we can solve these equations simultaneously to find critical points.
From the second equation, we can solve for y in terms of λ:
y = 1 / (8λ)
Substituting this into the equation of the ellipse and simplifying, we get:
x^2 + 4 / (64λ^2) = 1
64λ^2 * x^2 + 4 = 64λ^2
64λ^2 * x^2 = 64λ^2 - 4
x^2 = (64λ^2 - 4) / (64λ^2)
x^2 = 1 - 4 / (64λ^2)
x^2 = 1 - 1 / (16λ^2)
x^2 = (16λ^2 - 1) / (16λ^2)
Now, we can substitute this into the first equation and solve for x in terms of λ:
3 = 2λx
3 = 2λ * √((16λ^2 - 1) / (16λ^2))
3 = 2λ√(16λ^2 - 1) / (4λ)
3(4λ) = 2√(16λ^2 - 1)
12λ = √(16λ^2 - 1)
144λ^2 = 16λ^2 - 1
128λ^2 = -1
λ^2 = -1 / 128
No real solutions exist for λ^2 = -1 / 128, which means there are no critical points.
Since the function has no critical points on the ellipse, we need to consider the endpoints of the ellipse as well as the points where the ellipse intersects the x and y axes.
Let's find the maximum and minimum values by substituting these points into the function f(x, y) = 3x + y.
1. Endpoint (1, 0):
f(1, 0) = 3(1) + 0 = 3
2. Endpoint (-1, 0):
f(-1, 0) = 3(-1) + 0 = -3
3. Intersection points of the ellipse with the x-axis:
Substituting y = 0 into the equation of the ellipse:
x^2 + 4(0)^2 = 1
x^2 = 1
x = 1 or x = -1
Substituting these points into the function f(x, y):
f(1, 0) = 3(1) + 0 = 3
f(-1, 0) = 3(-1) + 0 = -3
4. Intersection points of the ellipse with the y-axis:
Substituting x = 0 into the equation of the ellipse:
0^2 + 4y^2 = 1
4y^2 = 1
y^2 = 1/4
y = 1/2 or y = -1/2
Substituting these points into the function f(x, y):
f(0, 1/2) = 3(0) + 1/2 = 1/2
f(0, -1/2) = 3(0) - 1/2 = -1/2
Now we can determine the maximum and minimum values:
Maximum value: 3
Minimum value: -3