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

[Reefy]

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.

 

[ehild]

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.

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

 

[Reefy]

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

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

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.

 

[Ray Vickson]

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.

 

[ehild]

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

That is right.

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

 

[Reefy]

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.