Math 241: Multivariable calculus, Lecture 15 Lagrange Multipliers. go.illinois.edu/math241fa17 Wednesday, October 4th, 2017 go.illinois.edu/math241fa17. Math 241: Problem of the day Problem: Find the absolute maximum and minimum value of the function f (x; y) = x3 − x + y 2x + y + 2 on the unit disk D = f(x; y) j x2 + y 2 ≤ 1g. What guarantees that the max and min exist? go.illinois.edu/math241fa17. rf (x; y) = h3x2 − 1 + y 2; 2yx + 1i −1 If this is the zero vector we get 2xy + 1 = 0, or y = 2x . (Note that x nor y can be zero, so dividing by x or y is no problem.) Substitute in the first component to get 1 1 3x2 − 1 + = 0 =) 3x4 − x2 + = 0: 4x2 4 2 2 1 Put z = x , get 3z − z + 4 = 0. The discriminant b2 − 4ac = 1 − 3 = −2, so there are NO critical points. So we need to look at the boundary! Solution Part 1: Critical points If f (x; y) = x3 − x + y 2x + y + 2 the gradient vector is go.illinois.edu/math241fa17. −1 If this is the zero vector we get 2xy + 1 = 0, or y = 2x . (Note that x nor y can be zero, so dividing by x or y is no problem.) Substitute in the first component to get 1 1 3x2 − 1 + = 0 =) 3x4 − x2 + = 0: 4x2 4 2 2 1 Put z = x , get 3z − z + 4 = 0. The discriminant b2 − 4ac = 1 − 3 = −2, so there are NO critical points. So we need to look at the boundary! Solution Part 1: Critical points If f (x; y) = x3 − x + y 2x + y + 2 the gradient vector is rf (x; y) = h3x2 − 1 + y 2; 2yx + 1i go.illinois.edu/math241fa17. (Note that x nor y can be zero, so dividing by x or y is no problem.) Substitute in the first component to get 1 1 3x2 − 1 + = 0 =) 3x4 − x2 + = 0: 4x2 4 2 2 1 Put z = x , get 3z − z + 4 = 0. The discriminant b2 − 4ac = 1 − 3 = −2, so there are NO critical points. So we need to look at the boundary! Solution Part 1: Critical points If f (x; y) = x3 − x + y 2x + y + 2 the gradient vector is rf (x; y) = h3x2 − 1 + y 2; 2yx + 1i −1 If this is the zero vector we get 2xy + 1 = 0, or y = 2x . go.illinois.edu/math241fa17. Substitute in the first component to get 1 1 3x2 − 1 + = 0 =) 3x4 − x2 + = 0: 4x2 4 2 2 1 Put z = x , get 3z − z + 4 = 0. The discriminant b2 − 4ac = 1 − 3 = −2, so there are NO critical points. So we need to look at the boundary! Solution Part 1: Critical points If f (x; y) = x3 − x + y 2x + y + 2 the gradient vector is rf (x; y) = h3x2 − 1 + y 2; 2yx + 1i −1 If this is the zero vector we get 2xy + 1 = 0, or y = 2x . (Note that x nor y can be zero, so dividing by x or y is no problem.) go.illinois.edu/math241fa17. 2 2 1 Put z = x , get 3z − z + 4 = 0. The discriminant b2 − 4ac = 1 − 3 = −2, so there are NO critical points. So we need to look at the boundary! Solution Part 1: Critical points If f (x; y) = x3 − x + y 2x + y + 2 the gradient vector is rf (x; y) = h3x2 − 1 + y 2; 2yx + 1i −1 If this is the zero vector we get 2xy + 1 = 0, or y = 2x . (Note that x nor y can be zero, so dividing by x or y is no problem.) Substitute in the first component to get 1 1 3x2 − 1 + = 0 =) 3x4 − x2 + = 0: 4x2 4 go.illinois.edu/math241fa17. The discriminant b2 − 4ac = 1 − 3 = −2, so there are NO critical points. So we need to look at the boundary! Solution Part 1: Critical points If f (x; y) = x3 − x + y 2x + y + 2 the gradient vector is rf (x; y) = h3x2 − 1 + y 2; 2yx + 1i −1 If this is the zero vector we get 2xy + 1 = 0, or y = 2x . (Note that x nor y can be zero, so dividing by x or y is no problem.) Substitute in the first component to get 1 1 3x2 − 1 + = 0 =) 3x4 − x2 + = 0: 4x2 4 2 2 1 Put z = x , get 3z − z + 4 = 0. go.illinois.edu/math241fa17. So we need to look at the boundary! Solution Part 1: Critical points If f (x; y) = x3 − x + y 2x + y + 2 the gradient vector is rf (x; y) = h3x2 − 1 + y 2; 2yx + 1i −1 If this is the zero vector we get 2xy + 1 = 0, or y = 2x . (Note that x nor y can be zero, so dividing by x or y is no problem.) Substitute in the first component to get 1 1 3x2 − 1 + = 0 =) 3x4 − x2 + = 0: 4x2 4 2 2 1 Put z = x , get 3z − z + 4 = 0. The discriminant b2 − 4ac = 1 − 3 = −2, so there are NO critical points. go.illinois.edu/math241fa17. Solution Part 1: Critical points If f (x; y) = x3 − x + y 2x + y + 2 the gradient vector is rf (x; y) = h3x2 − 1 + y 2; 2yx + 1i −1 If this is the zero vector we get 2xy + 1 = 0, or y = 2x . (Note that x nor y can be zero, so dividing by x or y is no problem.) Substitute in the first component to get 1 1 3x2 − 1 + = 0 =) 3x4 − x2 + = 0: 4x2 4 2 2 1 Put z = x , get 3z − z + 4 = 0. The discriminant b2 − 4ac = 1 − 3 = −2, so there are NO critical points. So we need to look at the boundary! go.illinois.edu/math241fa17. Look on the boundary for points where Du~f (a; b) = 0, for u~ tangent to the boundary. In other words we want rf (a; b) · u~ = 0. But we know that u~ satisfies rg(a; b) · u~ = 0. So we want rf (a; b) = λrg(a; b) (*) for some constant λ, called the Lagrange multiplier. At such points where (∗) holds, the level curves of f and of g are tangent. Lagrange multipliers Idea IDEA: Function f (x; y) is defined on a domain with boundary given by g(x; y) = 0. Want: absolute max/min of f on boundary. go.illinois.edu/math241fa17. In other words we want rf (a; b) · u~ = 0. But we know that u~ satisfies rg(a; b) · u~ = 0. So we want rf (a; b) = λrg(a; b) (*) for some constant λ, called the Lagrange multiplier. At such points where (∗) holds, the level curves of f and of g are tangent. Lagrange multipliers Idea IDEA: Function f (x; y) is defined on a domain with boundary given by g(x; y) = 0. Want: absolute max/min of f on boundary. Look on the boundary for points where Du~f (a; b) = 0, for u~ tangent to the boundary. go.illinois.edu/math241fa17. But we know that u~ satisfies rg(a; b) · u~ = 0. So we want rf (a; b) = λrg(a; b) (*) for some constant λ, called the Lagrange multiplier. At such points where (∗) holds, the level curves of f and of g are tangent. Lagrange multipliers Idea IDEA: Function f (x; y) is defined on a domain with boundary given by g(x; y) = 0. Want: absolute max/min of f on boundary. Look on the boundary for points where Du~f (a; b) = 0, for u~ tangent to the boundary. In other words we want rf (a; b) · u~ = 0. go.illinois.edu/math241fa17. So we want rf (a; b) = λrg(a; b) (*) for some constant λ, called the Lagrange multiplier. At such points where (∗) holds, the level curves of f and of g are tangent. Lagrange multipliers Idea IDEA: Function f (x; y) is defined on a domain with boundary given by g(x; y) = 0. Want: absolute max/min of f on boundary. Look on the boundary for points where Du~f (a; b) = 0, for u~ tangent to the boundary. In other words we want rf (a; b) · u~ = 0. But we know that u~ satisfies rg(a; b) · u~ = 0. go.illinois.edu/math241fa17. At such points where (∗) holds, the level curves of f and of g are tangent. Lagrange multipliers Idea IDEA: Function f (x; y) is defined on a domain with boundary given by g(x; y) = 0. Want: absolute max/min of f on boundary. Look on the boundary for points where Du~f (a; b) = 0, for u~ tangent to the boundary. In other words we want rf (a; b) · u~ = 0. But we know that u~ satisfies rg(a; b) · u~ = 0. So we want rf (a; b) = λrg(a; b) (*) for some constant λ, called the Lagrange multiplier. go.illinois.edu/math241fa17. Lagrange multipliers Idea IDEA: Function f (x; y) is defined on a domain with boundary given by g(x; y) = 0. Want: absolute max/min of f on boundary. Look on the boundary for points where Du~f (a; b) = 0, for u~ tangent to the boundary. In other words we want rf (a; b) · u~ = 0. But we know that u~ satisfies rg(a; b) · u~ = 0. So we want rf (a; b) = λrg(a; b) (*) for some constant λ, called the Lagrange multiplier. At such points where (∗) holds, the level curves of f and of g are tangent. go.illinois.edu/math241fa17. rf (x; y) = h3x2 − 1 + y 2; 2yx + 1i Boundary g(x; y) = x2 + y 2 = 1 rg = h2x; 2yi Then rf = λrg gives 3x2 − 1 + y 2 = λ2x; 2yx + 1 = λ2y Solution (0; 1) or (0; −1).Max: f (0; 1) = 3, Min: f (0; −1) = 1.