Maximisation and Minimisation Problem (1 Viewer)

[qqmore]

A point lies on the curve whose equation is x^4 + y^4 = 1. Prove that the distance of P from the origin is at most 2^(1/4).


Can anyone help me with this problem?
Thanks.

 

[doink]

find the distance in terms of x and y, differentiate implicitly, set to 0 find values of x and y then test to see if they are max, find max sub back into the original and theres your answer.

 

[Iruka]

Actually, you don't have to worry about implicit differentiation.

This curve is obviously symmetric in both x and y, so you really only have to figure out what is happening in the first quadrant.

Also, it is probably easier to maximize the square of the distance, rather than the distance itself.

So, if the point is (x,y), then the distance squared, which I shall call D, is given by
D = x^2 + y^2.

But since (x,y) lies on this curve (and we shall assume it is in the first quadrant), y^2 = sqrt(1-x^4),

so you just sub that in and go from there.

 

[qqmore]

Thanks a lot!
Just a quick question, would it be necessary to square root both sides of D = sqrt(1-x^4) + x^2 to find the distance (not squared) before differentiating or would it be possible to just do Dd / dx so that the maximum value of x gives the maximum value of D and hence distance (not squared)?...

 

[Pwnage101]

Calculus FTW

dind expression for distance (D) in terms of x and y via pytho, then change into one variabel (x OR y, doesnt matter - juts has to include only 1), then differentiate, finde when this = 0, and sub this bak in original formala to get otehr value (y value, if u put distance in terms of x only), then sub both these bak into distance equation, and just to show t is a MAX T.P. and not a MIN T.P., differentiate again, or show point just to left and just to right, ie dD/dx: (+) ---> 0---->(-)

in relation to ur Q, I DIDNT USE implicit diff., just diff. the square root sign adn all