Finding Global Extrema on Disc x^2 + y^2 ≤ 1 for f(x,y) = xy + 5y

[jonroberts74]

Homework Statement

find the global extrema on the disc $$x^2 + y^2 \le 1$$

given the function $$f(x,y)=xy+5y$$

The Attempt at a Solution

For the interior of the disc

$$\nabla f = <y,x+5>$$

the critical point is (0,-5)

for the boundary of the disc
using lagrange multipliers

$$\left\{\begin{array}{cc}y=\lambda 2x \\ x+5 = \lambda 2y \\ x^2+y^2 =1 \end{array}\right.$$

solving for lambda

$$\lambda = \frac{y}{2x}; \lambda = \frac{x+5}{2y}$$
$$\frac{y}{2x}=\frac{x+5}{2y} \Rightarrow y = \pm \sqrt{x(x+5)}$$
now,
$$x^2 + (\pm \sqrt{x(x+5)})^2 = 1 \Rightarrow x = \frac{1}{4}(\pm\sqrt{33}-5)$$
subbing the x value into
$$y = \pm \sqrt{x(x+5)} \Rightarrow \pm \sqrt{\frac{1}{8}(5 \sqrt{33} - 21)}$$

I know to test those critical points in the original function but before I go further I want to make sure I have done everything up to it correctly

 

[LCKurtz]

Homework Statement

find the global extrema on the disc $$x^2 + y^2 \le 1$$

given the function $$f(x,y)=xy+5y$$

The Attempt at a Solution

For the interior of the disc

$$\nabla f = <y,x+5>$$

the critical point is (0,-5)

for the boundary of the disc
using lagrange multipliers

$$\left\{\begin{array}{cc}y=\lambda 2x \\ x+5 = \lambda 2y \\ x^2+y^2 =1 \end{array}\right.$$

solving for lambda

$$\lambda = \frac{y}{2x}; \lambda = \frac{x+5}{2y}$$
$$\frac{y}{2x}=\frac{x+5}{2y} \Rightarrow y = \pm \sqrt{x(x+5)}$$
now,
$$x^2 + (\pm \sqrt{x(x+5)})^2 = 1 \Rightarrow x = \frac{1}{4}(\pm\sqrt{33}-5)$$
subbing the x value into
$$y = \pm \sqrt{x(x+5)} \Rightarrow \pm \sqrt{\frac{1}{8}(5 \sqrt{33} - 21)}$$

I know to test those critical points in the original function but before I go further I want to make sure I have done everything up to it correctly

I got the same numbers using a different method.

 

[Ray Vickson]

Homework Statement

find the global extrema on the disc $$x^2 + y^2 \le 1$$

given the function $$f(x,y)=xy+5y$$

The Attempt at a Solution

For the interior of the disc

$$\nabla f = <y,x+5>$$

the critical point is (0,-5)

for the boundary of the disc
using lagrange multipliers

$$\left\{\begin{array}{cc}y=\lambda 2x \\ x+5 = \lambda 2y \\ x^2+y^2 =1 \end{array}\right.$$

solving for lambda

$$\lambda = \frac{y}{2x}; \lambda = \frac{x+5}{2y}$$
$$\frac{y}{2x}=\frac{x+5}{2y} \Rightarrow y = \pm \sqrt{x(x+5)}$$
now,
$$x^2 + (\pm \sqrt{x(x+5)})^2 = 1 \Rightarrow x = \frac{1}{4}(\pm\sqrt{33}-5)$$
subbing the x value into
$$y = \pm \sqrt{x(x+5)} \Rightarrow \pm \sqrt{\frac{1}{8}(5 \sqrt{33} - 21)}$$

I know to test those critical points in the original function but before I go further I want to make sure I have done everything up to it correctly

Be careful: one of your $x =\frac{1}{4}(\sqrt{33}-5)$ and $x =\frac{1}{4}(-\sqrt{33}-5)$ is incorrect; do you see why?

It would have been less troublesome to use the Lagrangian stationary conditions to solve for $x,y$ as functions of $\lambda$, then use these expressions in the constraint to get a single equation for $\lambda$. The two roots for $\lambda$ correspond to the min and the max, and when you then use those values in the $x,y$ expressions you are done: no $\pm$ problems to worry about.

 

[jonroberts74]

I am not currently aware of Lagrangian stationary conditions. I will use my googles but I always welcome better ways to do a problems. especially ones with less than optimal results.

I'll check those x values in the morning.

EDIT:

I found something about creating a new function and taking partials

$$F(x,y,z,\lambda)=f(x,y,z) - \lambda(g(x,y,z)-k)$$

and take the partials for x,y,z,lambda

 

[Ray Vickson]

I am not currently aware of Lagrangian stationary conditions. I will use my googles but I always welcome better ways to do a problems. especially ones with less than optimal results.

I'll check those x values in the morning.

You are aware of the conditions: you used them! They are just the two equations $x+5 = 2 \lambda y$ and $y = 2 \lambda x$. These are the "stationarity" conditions for the Lagrangian function $L = f(x,y) - \lambda g(x,y)$ in the problem
$$\min f(x,y)\\<br /> \text{subject to } g(x,y) = 0$$

BTW: the Lagrangian $L$ plays an important role in second-order tests for maxima and minima, but more on that later.

 

[jonroberts74]

ah, okay thank you. I just hadn't heard that terminology used yet.

 

[jonroberts74]

I went through it again, got the same number. I am not seeing why one of the x values is incorrect

 

[Ray Vickson]

I went through it again, got the same number. I am not seeing why one of the x values is incorrect

Whenever you use nonlinear manipulations (squaring equations, and the like) you can introduce false solutions, that is, so-called solutions that do not satisfy the original equations you started with. Check both $x = (1/4)(\sqrt{33}-5)$ and $x = (1/4)(-\sqrt{33}-5)$. There is no way that one of them can possibly satisfy $x^2 + y^2 \leq 1$ for any real value of $y$. If you don't believe it, try evaluating them numerically.