Determine the points on the curve $$x^4+y^4=1$$ that are closest and furthest away from the origin. Explain why this corresponds to extremizing the function $f(x,y)=x^2+y^2$ under the condition $x^4+y^4=1$.
I don't understand this question at all and would be grateful if someone could provide some explanation to this!
Also when we are asked to extremize a function, given the constraint like in this example, how exactly do we find the function generally?
Hint: the function $f(x,y)$ represents the (squared of the) euclidean distance of a point $(x,y)$ from the origin.
Let's find the shortest and longest distances from the origin $O = (0,0)$ of points on the curve $x^4 + y^4 = 1$. The distance is $d = \sqrt{(x-0)^2 + (y-0)^2} = \sqrt{x^2 + y^2}$. So :$d^4 = (x^2 + y^2)^2 = (1\cdot x^2 + 1\cdot y^2)^2 \leq (1^2 + 1^2)\cdot ((x^2)^2 + (y^2)^2) = 2\cdot (x^4 + y^4) = 2\cdot 1 = 2$ due to Cauchy-Schwarz inequality. So $d_{max} = 2^{\frac{1}{4}} = 1.189$. This value is achieved when $(x,y) = (2^{-1/4}, 2^{-1/4}), (2^{-1/4},-2^{-1/4}),(-2^{-1/4},2^{-1/4}),(-2^{-1/4},-2^{-1/4})$
To find the shortest distance $d_{min}$, we observe that: if $x^4 + y^4 = 1$, then $|x| \leq 1$, and $|y| \leq 1$, implying that $x^2 \geq x^4$, and $y^2 \geq y^4$ with equality when $x = 0, 1$, and $y = 0,1$. So: $x^2 + y^2 \geq x^4 + y^4 = 1$. Thus: $d_{min} \geq \sqrt{1} = 1$, and this happens when $(x,y) = (0,1), (1,0),(-1,0),(0,-1)$
closest and furthest away from the origin– rubik May 04 '14 at 18:57