What are the steps for solving a problem using Lagrange Multipliers?

[harpazo]

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

Maximize: f (x, y) = e^(xy)

Constraint: x^2 + y^2 = 8

My Work:

I decided to rewrite the constraint as x^2 + y^2 without the constant 8 as originally given.

g (x, y) = x^2 + y^2

I found the gradient of f to be ye^(xy)i + xe^(xy)j.

I found the gradient of g to be 2xi + 2yj.

I then substituted the above into

gradient of f = L * gradient of g, where L represents the lowercase Greek letter lambda.

ye^(xy)i + xe^(xy)j = L * 2xi + 2yj.

I equated the coefficient of i to 2xL and the coefficient of j to 2yL.

This yields the following system of equations:

ye^(xy) = 2xL
xe^(xy) = 2yL

I am stuck here.

 

[MarkFL]

If you solve both equations for $\lambda$ and then equate the results, you obtain:

$$\frac{ye^{xy}}{2x}=\frac{xe^{xy}}{2y}$$

Multiply through by 2:

$$\frac{ye^{xy}}{x}=\frac{xe^{xy}}{y}$$

Since $e^{xy}\ne0$ for any real values, you may divide through by that expression:

$$\frac{y}{x}=\frac{x}{y}$$

What does this imply?

 

[harpazo]

If you solve both equations for $\lambda$ and then equate the results, you obtain:

$$\frac{ye^{xy}}{2x}=\frac{xe^{xy}}{2y}$$

Multiply through by 2:

$$\frac{ye^{xy}}{x}=\frac{xe^{xy}}{y}$$

Since $e^{xy}\ne0$ for any real values, you may divide through by that expression:

$$\frac{y}{x}=\frac{x}{y}$$

What does this imply?

1. Why can't e^(xy) = 0? What is the reason for this fact?

2. After dividing through using e^(xy), we are left with
y/x = x/y.

I then cross-multiply to find x^2 = y^2 but I do not know what the implication is in this case.

 

[MarkFL]

1. Why can't e^(xy) = 0? What is the reason for this fact?

Consider:

$$e^u=0$$

What do you get when solving for $u$?

2. After dividing through using e^(xy), we are left with
y/x = x/y.

I then cross-multiply to find x^2 = y^2 but I do not know what the implication is in this case.

Okay, you correctly found $x^2=y^2$...what do you get when you substitute for either $x$ or $y$ in the constraint?

 

[harpazo]

Consider:

$$e^u=0$$

What do you get when solving for $u$?
Okay, you correctly found $x^2=y^2$...what do you get when you substitute for either $x$ or $y$ in the constraint?

1. e^u = 0

Solving for u we get undefined.

2. I will plug either x^2 or y^2 into the constraint and get back to you later.

 

[harpazo]

Consider:

$$e^u=0$$

What do you get when solving for $u$?
Okay, you correctly found $x^2=y^2$...what do you get when you substitute for either $x$ or $y$ in the constraint?

Ok. I will take it from x^2 = y^2. I decided to plug x^2 for y^2 in the given constraint which is x^2 + y^2 = 8.

x^2 + x^2 = 8

After doing algebra, I found x to be -2 and 2.

I then substituted x = -2 & x = 2 into the constraint to find the y value(s).

I found the following 4 points which I denoted using A through D:

A (-2, 2)

B (-2, -2)

C (2, 2)

D (2, -2)

I rejected points A, B, and D because they fall outside of quadrant 1.

The critical point is (2, 2).

I finally substituted the critical point (2, 2) into the original function. I found the max value to be e^4 > 0. This max value happens at the critical point (2, 2).

What do you say?

 

[MarkFL]

I agree that the point $(2,2)$, is the only one that meets all criteria. Now we need to compare the value of $f$ at another point on the constraint, such as $(\sqrt{5},\sqrt{3})$, to determine if our critical point is a maximum or a minimum...what do you find?

 

[harpazo]

I agree that the point $(2,2)$, is the only one that meets all criteria. Now we need to compare the value of $f$ at another point on the constraint, such as $(\sqrt{5},\sqrt{3})$, to determine if our critical point is a maximum or a minimum...what do you find?

1. In Terms of the other point, where did it come from? Did it come from quadrant 1?

2. The back of the book tells me that I am correct. The max value is e^4 and it can only be found at the critical point (2, 2).

3. How can I solve this problem using Lagrange Multipliers? Like I said in another post, I enjoy working with Lambda.

 

[MarkFL]

I arbitrarily chose another point on the constraint, so that we could do a comparison like I mentioned just now in the other thread. :D

 

[harpazo]

I arbitrarily chose another point on the constraint, so that we could do a comparison like I mentioned just now in the other thread. :D

Good information. I have two more questions in terms of Lagrange Multipliers. The next chapter is DOUBLE INTEGRALS. Remember that I only present my questions after trying several times on my own.