calculus help asap

Question

For each of the following functions, find the maximum and mimimum values of the function on the circular disk x^2+y^2 =1

Do this by looking at the level curves and gradiants

Find the minimum and maximum value of the following

f(x,y)= x+y+1

f(x,y)= 1x^2+2y^2

f(x,)= 1x^2-2y^2

Answer 1

Any maximum/minimum value of function on a closed set is either

a critical point, or lies on the boundary.

To find maximum/minimum on the boundary we use the method of Lagrange

multipliers, which I assume is meant by "level curves and gradients".

1)

f(x,y) = x + y + 1

To find critical points:

∂f/∂x = 1

∂f/∂y = 1

Therefore there no critical points.

To find maximum/minimum along

x²+y²=1 (1)

we form

F(x,y,λ) = x + y + 1 - λ(x² + y² - 1)

∂F/∂x = 1 - 2λx = 0 ==> x = 1/2λ

∂F/∂y = 1 - 2λy = 0 ==> y = 1/2λ

Plugging it into the constraint (1)

1/4λ² + 1/4λ² = 1

λ = ±1/√2

This gives two points (x, y)

(√2/2, √2/2) and (-√2/2, -√2/2).

The first is global maximum and the second is global minimum.

2)

f(x,y) = x² + 2y²

To find critical points:

∂f/∂x = 2x

∂f/∂y = 4y

So we have one critical point (0,0).

In order to see if it is a local minimum, maximum, or saddle point,

we need the Hessian matrix H:

H = | 2 0 |

| 0 4 |

We see that its eigenvalues are positive (and so is the determinant).

Therefore the point (0,0) is a local minimum.

To compute the maximum/minimum along the boundary we again form

F(x,y,λ) = x² + 2y² - λ(x² + y² - 1)

∂F/∂x = 2x - 2λx = 0 ==> x(1 - λ) = 0

∂F/∂y = 4y - 2λy = 0 ==> y(2 - λ) = 0

The above two equations are satisfied if

- x = 0, in which case (1) implies y = ±1, which gives points (0, 1), (0, -1)

- y = 0, in which case (1) implies x = ±1, which gives points (1, 0), (1, 0)

- λ = 1, in which case y = 0 -- already covered above

- λ = 2, in which case x = 0 -- already covered above

So in total we have the following candidates and their corresponding values of f

f(0,0) = 0,

f(0,1) = 2,

f(0,-1) = 2,

f(1,0) = 1,

f(-1,0) = 1.

Therefore (0,0) is global minimum and both (0,1),(0,-1) are global maxima.

3)

f(x,y) = x² - 2y²

∂f/∂x = 2x

∂f/∂y = -4y

So we have one critical point (0,0).

In order to see if it is a local minimum, maximum, or saddle point,

we need the Hessian matrix H:

H = | 2 0 |

| 0 -4 |

Its determinant

D = ∂²f/∂x² ∂²f/∂y² - ∂²f/∂x∂y ∂²f/∂y∂x = -8 < 0

Therefore (0,0) is a saddle point.

To compute the maximum/minimum along he boundary we again form

F(x,y,λ) = x² - 2y² - λ(x² + y² - 1)

∂F/∂x = 2x - 2λx = 0 ==> x(1 - λ) = 0

∂F/∂y = -4y - 2λy = 0 ==> y(2 + λ) = 0

The above two equations are satisfied if

- x = 0, in which case (1) implies y = ±1, which gives points (0, 1), (0, -1)

- y = 0, in which case (1) implies x = ±1, which gives points (1, 0), (1, 0)

- λ = 1, in which case y = 0 -- already covered above

- λ = -2, in which case x = 0 -- already covered above

So in total we have the following candidates and their corresponding values of f

f(0,1) = -2,

f(0,-1) = -2,

f(1,0) = 1,

f(-1,0) = 1.

Therefore (0,1),(0,-1) are global minima and (1,0),(-1,0) are global maxima.