What are the absolute extrema of the given function on the given set?

In summary, the conversation discussed finding the absolute extrema of the function f(x,y)= x^2 + 4y^2 + 3x -1 on the set D={(x,y) | x^2 + y^2 ≤ 4}. The student initially found the critical point to be (-3/2,0) but later realized their mistake and found the correct critical point to be (-3/2, -13/4). The conversation also discussed finding the maximum value, which can also be found using the derivative and plugging in the critical point.
  • #1
Reefy
63
1

Homework Statement



Find the absolute extrema of the function on the set D.

f(x,y)= x^2 + 4y^2 + 3x -1

D= {(x,y) l x^2 + y^2 ≤ 4}

Homework Equations





The Attempt at a Solution



The only thing I've done so far was find the critical point. I found f-sub x = 2x+3 and f-suby= 8y giving me a critical point of (-3/2,0), correct?

Using that critical point, I found f(-3/2,0) = -1

After that, I don't know how to proceed. I know -2 ≤ x ≤ 2 and -2 ≤ y ≤ 2 but how do I find the other critical numbers with that info? I'm unsure because the closed boundary graph is a circle and not a square, box, or triangular region.
 
Physics news on Phys.org
  • #2
Reefy said:

Homework Statement



Find the absolute extrema of the function on the set D.

f(x,y)= x^2 + 4y^2 + 3x -1

D= {(x,y) l x^2 + y^2 ≤ 4}

Homework Equations


The Attempt at a Solution



The only thing I've done so far was find the critical point. I found f-sub x = 2x+3 and f-suby= 8y giving me a critical point of (-3/2,0), correct?

Using that critical point, I found f(-3/2,0) = -1

f(-3/2) is not equal to -1.

Reefy said:
After that, I don't know how to proceed. I know -2 ≤ x ≤ 2 and -2 ≤ y ≤ 2 but how do I find the other critical numbers with that info? I'm unsure because the closed boundary graph is a circle and not a square, box, or triangular region.

The absolute extreme can be on the perimeter. Find the critical point with the condition that x2+y2=4.

ehild
 
  • #3
ehild said:
f(-3/2) is not equal to -1.

Ah, you're right, thank you. For that, I got -13/4.

ehild said:
The absolute extreme can be on the perimeter. Find the critical point with the condition that x2+y2=4.

ehild

So set y=√(4-x^2) ? and find f(x,√(4-x^2)). Then find values of x and use the result to find y. Then find critical numbers? Because that's what I did but it didn't seem right.
 
Last edited:
  • #4
Reefy said:

Homework Statement



Find the absolute extrema of the function on the set D.

f(x,y)= x^2 + 4y^2 + 3x -1

D= {(x,y) l x^2 + y^2 ≤ 4}

Homework Equations





The Attempt at a Solution



The only thing I've done so far was find the critical point. I found f-sub x = 2x+3 and f-suby= 8y giving me a critical point of (-3/2,0), correct?

Using that critical point, I found f(-3/2,0) = -1

After that, I don't know how to proceed. I know -2 ≤ x ≤ 2 and -2 ≤ y ≤ 2 but how do I find the other critical numbers with that info? I'm unsure because the closed boundary graph is a circle and not a square, box, or triangular region.

You have found the global minimum of f, subject to g = x^2 + y^2 <= 4. (although your f is wrong while your (x,y) is correct). However, an extremum can also be a maximum, and you have not yet found that.
 
  • #5
Reefy said:
Ah, you're right, thank you. For that, I got -13/4.
That is right.

Reefy said:
So set y=√(4-x^2) ? and find f(x,√(4-x^2)). Then find values of x and use the result to find y. Then find critical numbers? Because that's what I did but it didn't seem right.

Show your work.

ehild
 
  • #6
Thanks, guys! I got it now. I forgot to take the derivative of f(x,√(4-x^2)).

Instead of finding the derivative, I found x and y of f(x,√(4-x^2)) and plugged it into the original function which was wrong.
 

FAQ: What are the absolute extrema of the given function on the given set?

1. What is the definition of absolute extrema?

Absolute extrema refer to the highest and lowest values of a function within a given interval. These values are also known as the maximum and minimum values.

2. How do you find the absolute extrema of a function with 2 variables?

To find the absolute extrema of a function with 2 variables, you first need to determine the critical points by taking the partial derivatives of the function with respect to each variable and setting them equal to 0. Then, plug these critical points into the original function to determine which one gives the highest and lowest values.

3. Can a function have more than one absolute extrema?

Yes, a function can have multiple absolute extrema within a given interval. These extrema can be either local (within a smaller interval) or global (within the entire interval).

4. How do you know if a critical point is a maximum or minimum?

To determine if a critical point is a maximum or minimum, you can use the second derivative test. If the second derivative is positive at the critical point, it is a minimum. If the second derivative is negative, it is a maximum. If the second derivative is 0, the test is inconclusive.

5. Can a function have absolute extrema at the endpoints of an interval?

Yes, a function can have absolute extrema at the endpoints of an interval. This is because the endpoints are also considered critical points, and they can give the highest or lowest values of the function within the given interval.

Similar threads

Back
Top