Finding Extrema with Lagrange Multipliers

[harpazo]

Use Lagrange Multipliers to find the individual extrema, assuming that x and y are positive.

Maximize: f (x, y) = sqrt {6 - x^2 - y^2}

Constraint: x + y - 2 = 0

My Work:

I first decided to rewrite the constraint as g (x, y) = x + y without the constant -2 as originally given.

I found the gradient of f to be

(-xi - yj)/sqrt {6 - x^2 - y^2}.

I found the gradient of g to be i + j.

Let L = the lowercase Greek letter lambda for short.

I substituted the gradient of f and g into

gradient of f = L * gradient of g.

[-x/sqrt {6 - x^2 - y^2}]i - [y/sqrt {6 - x^2 - y^2}]j = L (i + j)

I equated the coefficient of i to L and the coefficient of j to L.

-x/sqrt {6 - x^2 - y^2} = L

-y/sqrt {6 - x^2 - y^2} = L

I am struck here.

 

[MarkFL]

Okay, so what this implies is:

$$\frac{x}{\sqrt{6-x^2-y^2}}=\frac{y}{\sqrt{6-x^2-y^2}}$$

Cross-multiply:

$$x\sqrt{6-x^2-y^2}=y\sqrt{6-x^2-y^2}$$

$$x\sqrt{6-x^2-y^2}-y\sqrt{6-x^2-y^2}=0$$

$$(x-y)\sqrt{6-x^2-y^2}=0$$

So, what are your critical point(s)?

 

[harpazo]

Okay, so what this implies is:

$$\frac{x}{\sqrt{6-x^2-y^2}}=\frac{y}{\sqrt{6-x^2-y^2}}$$

Cross-multiply:

$$x\sqrt{6-x^2-y^2}=y\sqrt{6-x^2-y^2}$$

$$x\sqrt{6-x^2-y^2}-y\sqrt{6-x^2-y^2}=0$$

$$(x-y)\sqrt{6-x^2-y^2}=0$$

So, what are your critical point(s)?

To find the critical points, I set (x - y) = 0 and
sqrt {6 - x^2 -y^2} = 0 and then solve for x and y.
Let me know if this works for locating the critical points.

 

[MarkFL]

To find the critical points, I set (x - y) = 0 and
sqrt {6 - x^2 -y^2} = 0 and then solve for x and y.
Let me know if this works for locating the critical points.

What I would do is use the constraint to determine $y=2-x$. Now substitute for $y$ in both equations you mentioned, and solve for $x$, then your critical points will be $(x,2-x)$, and you will want to discard any points not in the first quadrant, since you stated both $x$ and $y$ are positive. What do you find?

 

[harpazo]

What I would do is use the constraint to determine $y=2-x$. Now substitute for $y$ in both equations you mentioned, and solve for $x$, then your critical points will be $(x,2-x)$, and you will want to discard any points not in the first quadrant, since you stated both $x$ and $y$ are positive. What do you find?

I will do as you suggested. What do I find? Are we not seeking critical points? I will work on this later and get back to you.

 

[harpazo]

What I would do is use the constraint to determine $y=2-x$. Now substitute for $y$ in both equations you mentioned, and solve for $x$, then your critical points will be $(x,2-x)$, and you will want to discard any points not in the first quadrant, since you stated both $x$ and $y$ are positive. What do you find?

Hello Mark. I got home about 20 minutes ago and decided to head straight to this question. Math helps me forget my problems in life. It may sound silly but it's true.

I did what you suggested. The only critical point that applies here and lies in quadrant 1 is (1, 1).

I then evaluated the given function f(1, 1). The max value is 2 and it happens at the critical point (1, 1).

What do you think?

 

[MarkFL]

Hello Mark. I got home about 20 minutes ago and decided to head straight to this question. Math helps me forget my problems in life. It may sound silly but it's true.

I did what you suggested. The only critical point that applies here and lies in quadrant 1 is (1, 1).

I then evaluated the given function f(1, 1). The max value is 2 and it happens at the critical point (1, 1).

What do you think?

I agree that of the 3 critical points, $(1,1)$ is the only one in quadrant I. Now, we know this is either a maximum or a minimum, and to determine which, we can choose another point on the constraint, such as $$\left(\frac{3}{2},\frac{1}{2}\right)$$. We then find:

$$f\left(\frac{3}{2},\frac{1}{2}\right)=\sqrt{\frac{7}{2}}<2$$

So now we may assert that:

$$f_{\max}=f(1,1)=2$$

 

[harpazo]

I agree that of the 3 critical points, $(1,1)$ is the only one in quadrant I. Now, we know this is either a maximum or a minimum, and to determine which, we can choose another point on the constraint, such as $$\left(\frac{3}{2},\frac{1}{2}\right)$$. We then find:

$$f\left(\frac{3}{2},\frac{1}{2}\right)=\sqrt{\frac{7}{2}}<2$$

So now we may assert that:

$$f_{\max}=f(1,1)=2$$

1. Where did the other point come from?

2. How can I solve this problem via Lagrange Multipliers?
I enjoy using Lambda.

3. The answer in the back of the book reveals the fact that we are correct.

 

[MarkFL]

1. Where did the other point come from?

2. How can I solve this problem via Lagrange Multipliers?
I enjoy using Lambda.

3. The answer in the back of the book reveals the fact that we are correct.

I chose the point as it is on the constraint. Using that point, we can determine if our one critical point is a maximum or a minimum. If the objective function evaluated at the other point is above the value of the objective function at the critical point, then we know the critical point is at the minimum, likewise if the objective function evaluated at the other point is below the value of the objective function at the critical point, then we know the critical point is at the maximum.

 

[harpazo]

I chose the point as it is on the constraint. Using that point, we can determine if our one critical point is a maximum or a minimum. If the objective function evaluated at the other point is above the value of the objective function at the critical point, then we know the critical point is at the minimum, likewise if the objective function evaluated at the other point is below the value of the objective function at the critical point, then we know the critical point is at the maximum.

Good information here.