Finding Max/Min Values on Functions of 2 Variables

I'm just beginning to think that I don't have a solution to this problem. Homework Statement Let g(x,y) = x2 + 4y2. What is the maximum value of F(x,y) = ln(x4y5) on the intersection of the level set g(x,y) = 9 with the quadrant {(x,y): x>0 and y>0}The Attempt at a SolutionIn summary, the maximum value of F(x,y) = ln(x4y5) on the intersection of the level set g(x,y) = 9 with the quadrant {(x,y): x>0 and y>0} cannot be found as it results in a singularity at the boundary of
  • #1

Homework Statement


Let g(x,y) = x2 + 4y2. What is the maximum value of F(x,y) = ln(x4y5) on the intersection of the level set g(x,y) = 9 with the quadrant {(x,y): x>0 and y>0}

The Attempt at a Solution


It seems I'm having a lot of difficulty with Lagrange multipliers, but here I go.

Fx = 4/x = gx = 2x * λ
Fy = 5/y = gy = 8y * λ

Then clearly, x2 = 2/λ
and, y2 = 5/(8λ)

Plugging these into the constraint gives,

g(2/λ,5/(8λ)) = 4/λ2 + 4 * (25/(64 * λ2)) = 9.

Attempting to solve for λ gives,

(4/λ2) (1 + 25/64) = 9
4 + 25/16 = 9λ2
λ = ± √(4/9 + 25/144). But I reject the negative because it will yield answers outside the boundary

x = sqrt(2/λ) = sqrt(2/(sqrt((4/9 + 25/144))) ≈ 1.594990568
y = sqrt(5/(8λ) = sqrt(5/(8sqrt((4/9 + 25/144))) ≈ 0.89162683338

So I get an extremum at F(1.594990568,0.89162683338), whatever that is...

Pretty sure this isn't right. Ideas?
 
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  • #2
Hi TranscendArcu! :smile:

When you plugged your values into the constraint, you took the square again, when you already had a square...
 
  • #3
I find that it is often simplest to eliminate [itex]\lambda[/itex] (which is not necessary for the solultion) first by dividing one equation by another.\
Here, you have [itex]4/x= 2\lambda x[/itex] and [itex]5/y= 8\lambda y=[/itex]. dividing the first by the second gives (4/5)(y/x)= x/(4y) or 16y^2= 5x^2. [itex]y= \pm (\
sqrt{5}/4)x[/itex]. Substitute that into the constraint and solve for x.
 
  • #4
However, given the constraint y > 0, we would reject the negative answer for y = (sqrt5/4)x, right?

Supposing I solved out for x instead of y. I begin with 4y/(5x) = x/(4y), which gives 16y^2 = 5x^2. So,

x = 4y/sqrt(5)

Subbing into x^2 + 4y^2 = 9 gives (16y^2/5) + 20y^2/5 = 9. So, 36y^2 = 45, or

y = sqrt(45)/6 = sqrt(5)/2

By our definition of x in terms of y: x = 4(sqrt(5)/2)/sqrt(5) = 2.

Therefore, our maximum value is ln(16 * 5^(5/2)/4^5) = ln(5^(5/2)/64) and exists at (2,sqrt(5)/2)

Sound about right?
 
  • #5
Yes, that sounds right.

Btw, with a problem like this, you should also pay attention to x=0 and y=0 that might have boundary extrema.
 
  • #6
But given the constraint (x,y): x>0 and y>0, shouldn't be the cases of x=0 and y=0 be non-issues?
 
  • #7
TranscendArcu said:
But given the constraint (x,y): x>0 and y>0, shouldn't be the cases of x=0 and y=0 be non-issues?

Suppose that at x=0.0001 your function would have a higher value than your "maximum"?
 
  • #8
TranscendArcu said:

Homework Statement


Let g(x,y) = x2 + 4y2. What is the maximum value of F(x,y) = ln(x4y5) on the intersection of the level set g(x,y) = 9 with the quadrant {(x,y): x>0 and y>0}

The Attempt at a Solution


It seems I'm having a lot of difficulty with Lagrange multipliers, but here I go.

Fx = 4/x = gx = 2x * λ
Fy = 5/y = gy = 8y * λ

Then clearly, x2 = 2/λ
and, y2 = 5/(8λ)

Plugging these into the constraint gives,

g(2/λ,5/(8λ)) = 4/λ2 + 4 * (25/(64 * λ2)) = 9.

Attempting to solve for λ gives,

(4/λ2) (1 + 25/64) = 9
4 + 25/16 = 9λ2
λ = ± √(4/9 + 25/144). But I reject the negative because it will yield answers outside the boundary

x = sqrt(2/λ) = sqrt(2/(sqrt((4/9 + 25/144))) ≈ 1.594990568
y = sqrt(5/(8λ) = sqrt(5/(8sqrt((4/9 + 25/144))) ≈ 0.89162683338

So I get an extremum at F(1.594990568,0.89162683338), whatever that is...

Pretty sure this isn't right. Ideas?

[itex] F_x [/itex] is not equal to [itex] 4/x [/itex] , and [itex] F_x [/itex] is not equal to [itex] g_x [/itex], although later you set it equal to [itex] \lambda g_x, [/itex] which is OK. Start again, with correct derivatives for F, and be sure to write the Lagrange equations correctly.

RGV
 
  • #9
I like Serena said:
Suppose that at x=0.0001 your function would have a higher value than your "maximum"?

Alright, so let me test the boundaries of the quadrant:

Suppose x = 0 and y ≥ 0, then ln(x4y5) goes to -∞. Right?

Suppose y = 0 and x ≥ 0, then ln(x4y5) also goes to -∞.

Hmm... Maybe I just don't know how to check a boundary. How do you check a boundary?
 
  • #10
Ray Vickson said:
[itex] F_x [/itex] is not equal to [itex] 4/x [/itex] , and [itex] F_x [/itex] is not equal to [itex] g_x [/itex], although later you set it equal to [itex] \lambda g_x, [/itex] which is OK. Start again, with correct derivatives for F, and be sure to write the Lagrange equations correctly.

RGV
I checked Fx with Wolfram Alpha: http://www.wolframalpha.com/input/?i=d/dx+ln(x^4y^5)

What am I doing wrong?
 
  • #11
TranscendArcu said:
Alright, so let me test the boundaries of the quadrant:

Suppose x = 0 and y ≥ 0, then ln(x4y5) goes to -∞. Right?

Suppose y = 0 and x ≥ 0, then ln(x4y5) also goes to -∞.

Hmm... Maybe I just don't know how to check a boundary. How do you check a boundary?

You just did.
Note that your function won't have a global minimum, since for x or y close to zero it approaches minus infinity.
But the extremum that you did find, has to be a maximum now.
 
  • #12
TranscendArcu said:
I checked Fx with Wolfram Alpha: http://www.wolframalpha.com/input/?i=d/dx+ln(x^4y^5)

What am I doing wrong?

Sorry: your F_x is correct. I was thinking about using x^4*y^5, rather than its logarithm.

Anyway, your major error (besides saying F_x = g_x---which is FALSE) is that you have x^2 = 2/lambda, etc., so when you substitute these into g you should NOT get 1/lambda^2. You are substituting x^4 and y^4 into g, which is incorrect.

RGV
 

What is the definition of a max/min value on a function of 2 variables?

A max/min value on a function of 2 variables is the highest or lowest point on the graph of the function where the output (dependent variable) takes its maximum or minimum value for a given input (independent variable).

How do you find the max/min values of a function of 2 variables?

To find the max/min values of a function of 2 variables, you can use the partial derivative test. This involves finding the first and second partial derivatives of the function with respect to each variable, setting them equal to zero, and solving for the critical points. The critical points can then be evaluated to determine if they are max/min values.

What is the significance of finding the max/min values of a function of 2 variables?

Finding the max/min values of a function of 2 variables is important for understanding the behavior of the function and determining its optimal points. This information can be useful in various fields such as economics, physics, and engineering where optimizing a function is crucial.

Can a function of 2 variables have more than one max/min value?

Yes, a function of 2 variables can have multiple max/min values. These points are known as local maxima/minima and can occur at different locations on the graph of the function.

How does the method for finding max/min values on a function of 2 variables differ from that of a function of 1 variable?

The method for finding max/min values on a function of 2 variables is more complex compared to a function of 1 variable. This is because there are multiple directions in which the function can change and thus, multiple critical points to consider. In contrast, a function of 1 variable only has one critical point to evaluate.

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