MHB Optimization Problem: Max and Min Values of e^(x+2y) with x^2+y^2=5 Condition

[Yankel]

Hello all,

I am trying to solve this problem, and I don't know which way to go...

The product of the maximum value and minimum value of the function:

\[z=e^{x+2y}\]

under the condition:

\[x^{2}+y^{2}=5\]

Equals to:

a. 0
b. 1
c. e^5
d. e
e. e^-1

I have a feeling that this problem involves the Lagrange multipliers, but I am really not sure, and doesn't know how it's related to it (in case I am correct).

Thank you

 

[MarkFL]

Yes, Lagrange multipliers is a good way to go here. Your objective function is:

$$f(x,y)=e^{x+2y}$$

subject to the constraint:

$$g(x,y)=x^2+y^2-5=0$$

Can you form the system of equations that results from the partials?

 

[I like Serena]

Hi!

Let's indeed try with the Lagrange multipliers.

We define:
$$\Lambda = e^{x+2y} - \lambda(x^{2}+y^{2}-5)$$
And we set each of the partial derivatives of $\Lambda$ to zero.
Which system do you get?

 

[Yankel]

I did what you suggested and got these 3 equations, technically (algebraically), how do I solve such a thing ?\[e^{x+2y}+2\lambda x=0\]

\[2e^{x+2y}+2\lambda y=0\]

\[x^{2}+y^{2}-5=0\]Thank you

 

[MarkFL]

You have a sign error in front of the terms having $\lambda$ as a factor.

The way I was taught, which is equivalent to I like Serena's suggestion, is to use:

$$f_x(x,y)=\lambda g_x(x,y)$$

$$f_y(x,y)=\lambda g_y(x,y)$$

Now this should give you an implication regarding the relationship between the two variables at the critical point. Solve both for $\lambda$ and then equate the results and simplify. What do you find?

 

[Yankel]

I can't find any error. The Lagrangian function is:

\[L(x,y,\lambda )=e^{x+2y}+\lambda (x^{2}+y^{2}-5)\]

And I need to find it's derivatives by x, y and lambda, can't find my error there.

 

[MarkFL]

I can't find any error. The Lagrangian function is:

\[L(x,y,\lambda )=e^{x+2y}+\lambda (x^{2}+y^{2}-5)\]

And I need to find it's derivatives by x, y and lambda, can't find my error there.

Well, you are correct in that within the Lagrangian function, the term with $\lambda$ may be either added or subtracted. I prefer subtraction as given by I like Serena. Either way, you should find the same implication though.

 

[Yankel]

I am getting confused here. I learned that I need to add, not subtract. I believe you that you can do both.

But assume that adding is what needed, do we agree that my derivatives are correct ?
(they must be, checked with Maple)

How do I solve this system then ?

 

[MarkFL]

I am getting confused here. I learned that I need to add, not subtract. I believe you that you can do both.

But assume that adding is what needed, do we agree that my derivatives are correct ?
(they must be, checked with Maple)

How do I solve this system then ?

Yes, my apologies...what you have done so far is correct. So, let's look at your first two equations:

$$e^{x+2y}+2\lambda x=0$$

$$2e^{x+2y}+2\lambda y=0$$

Now, solve both for $\lambda$ and then equate the results and simplify. What do you find?

 

[Yankel]

No worries :)

What I got was:

\[\frac{1}{2x}=\frac{1}{y}\]

From which I got:

\[y=2x\]

I put it in the 3rd equation and got:

\[x=\pm 1\rightarrow y=\pm 2\]

Leading me to 2 critical points:

\[f(1,2)=e^{5}\]

\[f(-1,-2)=e^{-5}\]Are my calculations correct ?

Are these the only points, how do I know that they are absolute min/max ?

 

[MarkFL]

Yes, your calculations are correct. Does the constraint lead to any other critical points with the implication you found? To verify the nature of the critical points you found, you could use the constraint, solving it for one of the variables, and then substitute into the objective function to get a function in one variable, and then use the first or second derivative test.

 

[I like Serena]

Are my calculations correct ?

Are these the only points, how do I know that they are absolute min/max ?

Yep. All correct!

Since you have a continuous function defined on a circle, there can be no boundary extremum.
That leaves only local and absolute minima and maxima.
Since there is only one minimum and one maximum, both of them have to be absolute.

 

[Yankel]

Thanks guys,

great help !

(Yes)