Solving Max of x^2+y^2 w/ Lagrange Multipliers

AI Thread Summary
To find the shortest and longest distance from the origin to the curve defined by x^2 + xy + y^2 = 16, the method of Lagrange multipliers is suggested to maximize x^2 + y^2. Participants express confusion about applying this method and suggest alternative approaches, including solving for y in terms of x and using the distance formula. The distance formula D = (x^2 + y^2)^(1/2) is noted as similar to the expression being maximized. The discussion emphasizes the need to derive the distance function, take its derivative, and find critical points for analysis. Overall, the conversation revolves around clarifying the application of calculus techniques to solve the problem effectively.
thenewbosco
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Find the shortest and longest distance from the origin to the curve
x^2 + xy + y^2=16 and give a geometric interpretation...the hint given is to find the maximum of x^2+y^2

i am not sure what to do for this problem

thanks
 
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thenewbosco said:
Find the shortest and longest distance from the origin to the curve
x^2 + xy + y^2=16 and give a geometric interpretation...the hint given is to find the maximum of x^2+y^2

i am not sure what to do for this problem

thanks

Are you sure you need Lagrange Multipliers for this?
 
it says for the hint to use the method of lagrange multipliers to find the maximum of x^2 + y^2 but i am not sure how to do it using any method, so any help is appreciated.
 
thenewbosco said:
it says for the hint to use the method of lagrange multipliers to find the maximum of x^2 + y^2 but i am not sure how to do it using any method, so any help is appreciated.

Solve for y. use rate of change respect to the distance.
that is the "cal 1 method"

the path equation is constraint i think. apply Lagrange Multipliers on the distance formula
 
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solve for y in what though. in the question it says x^2+y^2 this isn't even an equation though.
I am sorry i still don't get it
 
thenewbosco said:
solve for y in what though. in the question it says x^2+y^2 this isn't even an equation though.
I am sorry i still don't get it

you can solve for y in tern of x
and then using the distance formula D = (y^2+x^2)^0.5
sub the y equation into the distance formula
take the first derivative
fine 0s
test it
done

that is cal 1 method, it requires a lot of work

x^2+y^2 looks really similar to the distance formula
D^2 = x^2 + y^2

you can set D = f(x) or D^2 = f(x) and find the del of it, since the square doesn't where the extreme occurs, therefore the text tells you to fine the max of x^2+y^2
 
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