Find the maximum and minimum values of f(x,y)=6x+y on the ellipse x^2+49y^2=1

Answer 1

Our objective function is f(x,y)=6x+y given the constraint function g(x,y) = x²+49y²-1 = 0.

By creating the Lagrangian function, we have:

L(x,y,λ) = f(x,y) - λg(x,y) = (6x+y) - λ(x²+49y²-1) = 6x + y - x²λ - 49y²λ + λ

Next, we find the partial derivatives for variables x and y and set them equal to 0:

∂L/∂x = 6 - 2λx = 0 --> 6 = 2λx --> 3/λ = x

∂L/∂y = 1 - 98λy = 0 --> 1 = 98λy --> 1/(98λ) = y

Now we plug our new values for x and y and solve for lemma (λ):

x²+49y²-1=0

(3/λ)² + 49(1/(98λ))² - 1 = 0

9/λ² + 49/(9604λ²) - 1 = 0

9/λ² + 1/(196λ²) = 1

1764 = 196λ²

9 = λ²

±3 = λ

This means the possible values for x and y are x=±1 and y=±1/294. By plugging these values into the objective function, we can determine the maximum and minimum values:

f(1,1/294) = 6(1) + 1/294 = 1765/294 <-- Maximum

f(-1,1/294) = 6(-1) + 1/294 = -1763/294

f(1,-1/294) = 6(1) + (-1/294) = 1763/294

f(-1,-1/294) = 6(-1) + (-1/294) = -1765/294 <-- Minimum

Hence, the maximum and minimum values of f(x,y)=6x+y on the ellipse x²+49y²=1 are 1765/294 and -1765/294 respectively.

Hope this helped!

Answer 2

The standard is to use Lagrange multipliers (you can see the Wikipedia article for more information).

Letting L = (6x+ y) + c(x^2 + 49y^2-1) be our lagrangian function (notice the added constant c). We now want where the partial derivatives with respect to all the variables and c are 0. this gives us 3 equations

dL/dx = 6 + 2xc = 0

dL/dy = 1 + 98yc = 0

dL/dc = (x^2 + 49y^2-1) = 0

It's usually easier to solve for c first. The first equation gives x = -3/c and the second equation gives y = -1/98 c. Then putting this into our last equation (which is basically the same as the constraint equation we get)

9/c² + 1/(196 c²) = 1

We can now solve for c, and once we do that we will have x and y as well.