Extrema of multivariable function

In summary, the expert is trying to find the extrema of a function f(x,y) = (4x^2)(e^y) - 2x^4 - e^4y but is having difficulty because of the exponentials in the equation. To simplify the equation, the expert sets e^y-1 to zero and gets 1 and -1 which yields the points (-1,0) and (1,0). If x=0 then the other equation doesn't work.
  • #1
haroldholt
21
0
Hi

I'm studying for a calculus exam and I'm a little stuck on finding the extrema for multivariable functions.

For the particular question I'm trying to do now I need to find and classify the extrema for the function f(x,y) = (4x^2)(e^y) - 2x^4 - e^4y.

I can find the first derivatives, being fx = 8xe^y - 8x and fy = (4x^2)(e^y) - 4e^4y and I know I have to let them be equal to 0 to find where the extrema are located but I'm not sure how to do that. I guess it's just the exponentials that are throwing me off.

Any help would be appreciated.
 
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  • #2
You have done well so far. The next step is to simultaneously set those partial derivatives to zero. So try that and show us the results.
 
  • #3
Thanks

Well it's 0 = 8xe^y - 8x and 0 = (4x^2)(e^y) - 4e^4y. But I'm not sure where to go from there. I'm not sure what to do with the exponentials, I know they can never equal zero, but I'm not sure what that means for my equations.
 
  • #4
Simplify the fx=0 equation.
 
  • #5
0 = 8x(e^y - 1). But I still don't know what to do with the exponential.
 
  • #6
Any equation of the form a*b=0 tells you that either a=0 or b=0 (or both). This is obviously of that form. x=0 doesn't lead to any extrema. (Why not?) What happens when you set e^y-1 to zero?
 
  • #7
Thanks for that. I now realize that y can be set to zero in the equation 8x(e^y - 1). And then if you sub y = 0 into the other equation you get 1 and -1 which yields the points (-1,0) and (1,0) (which is the answer in the back of the book :smile:). And if x = 0 then the other equation doesn't work. Thanks heaps. I now see how stupid I was initially lol.
 
  • #8
I wish when I learned calculus (or any other courses) I could have such instruction.
 

Related to Extrema of multivariable function

1. What is an extrema of a multivariable function?

An extrema, or extreme value, of a multivariable function is a point where the function reaches either its maximum or minimum value. This can occur in multiple variables, such as x and y, and is often referred to as a local or global extrema.

2. How do you find the extrema of a multivariable function?

To find the extrema of a multivariable function, you must first take the partial derivatives of the function with respect to each variable. Then, set these derivatives equal to 0 and solve for the variables. The resulting points are potential extrema, and can be further analyzed using the second derivative test to determine if they are maximum or minimum values.

3. What is the second derivative test?

The second derivative test is a method used to determine whether a critical point of a multivariable function is a maximum or minimum value. It involves taking the second derivative of the function and evaluating it at the critical point. If the second derivative is positive, the critical point is a minimum, and if it is negative, the critical point is a maximum. If the second derivative is 0, further analysis is needed.

4. Can a multivariable function have multiple extrema?

Yes, a multivariable function can have multiple extrema. These can be local extrema, where the function reaches its maximum or minimum value at a specific point, or global extrema, where the function reaches its maximum or minimum value over the entire domain of the function.

5. How are extrema of multivariable functions used in real-world applications?

Extrema of multivariable functions are used in a variety of real-world applications, such as optimization problems in engineering and economics, and in determining the maximum or minimum values of physical quantities, such as temperature or pressure, in scientific experiments. They are also used in machine learning and data analysis to find the most efficient or optimal solutions.

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