Direct/Lagrange to show max/min

Question

The domain Is defined by the half unit circle: $\left \{ \left ( x,y \right )\in \mathbb{R}^2: x^2+y^2\leq 1, x\geq 0 \right \}$

for the function $f:D\rightarrow \mathbb{R}$. The function is $f(x,y)=2xy^2-2x^2$. I have shown that a min/max must exists since the function is continuous and the domain is both bounded and closed. I do not know how to proceed to find these points where the function is equal to its min/max. I tried with the Lagrange multiplier but it is just too messy for me to do. How would one do it directly? I found that the gradient vanishes at $(0,0)$

Answer 1

Since $x\geqslant 0, f(x,y) \leqslant 2x(1-x^2)-2x^2$. Now find $x$ between 0 and 1 to maximize this.

Finding the minimum is a lot easier since $2xy^2\geqslant 0$ so we must have $y=0$. Can you do the rest?

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We can also use polar coordinates. Let $x=R\cos \alpha, y=R\sin \alpha$, with $0\leqslant R \leqslant 1, -\frac{\pi}{2} \leqslant \alpha \leqslant \frac{\pi}{2}$.

(1) On boundary $S_1 = \left\{ x^2+y^2=1, x\geqslant 0 \right\}$ we have $R^2=1$ so $f(x,y) = 2\cos \alpha \sin^2 \alpha - 2 \cos^2 \alpha = 2\cos \alpha - 2\cos^3 \alpha - 2 \cos^2 \alpha$.

Now $\frac{\partial f}{\partial \alpha}= -2 \sin \alpha-6\cos^2\alpha (-\sin \alpha)-4 \cos \alpha (-\sin \alpha) \\ = 2 \sin \alpha (-1+3 \cos^2\alpha +2\cos \alpha)\\ = 2\sin\alpha (3\cos \alpha -1) (\cos \alpha+1) $

$\frac{\partial{f}}{\alpha}=0 \Rightarrow \sin \alpha =0 $ or $\cos \alpha = \frac{1}{3}$

When $R=1, \sin \alpha =0, x=1, y=0, f(x,y) = -1$.

When $R=1, \cos \alpha = \frac{1}{3}, x=\frac{1}{3}, y=\pm \frac{2\sqrt{2}}{3}, f(x,y)=\frac{10}{27}.$

(2) On boundary $S_2 = \left\{ x=0, -1\leqslant y \leqslant 1\right\}$ we have $f(x,y)=0$.

Therefore the minimum is $-1$ at $(1,0)$ and the maximum is $\frac{10}{27}$ at $(\frac{1}{3}, \pm \frac{2\sqrt{2}}{3})$

Answer 2

We may start by saying something about the function $ \ f(x,y) \ = \ 2xy^2 \ - \ 2x^2 \ = \ 2x · (y^2 - x) \ . $ Because the coordinate variable $ \ y \ $ appears in only one term with an even exponent, the value of the function is unaltered by a sign-change of $ \ y \ $; in other words, the function has symmetry about the $ \ x-$ axis. There is no symmetry for the variable $ \ x \ $ since it appears with both even and odd exponents. Further, we can say that $ \ f(0,y) \ = \ 0 \ $ and $ \ f(y^2,y) \ = \ 0 \ . $ This divides the plane into four regions, with $ f(x,y) \ < \ 0 \ $ for $ \ x \ < \ 0 \ $ and $ \ x \ > \ y^2 \ $ , and $ f(x,y) \ > \ 0 \ $ otherwise. The graph below presents the situation, with the unit circle marked in red; this will be useful in the discussion to come.

You found from the first partial derivative equations $ \ f_x \ = \ 2y^2 \ - \ 4x \ = \ 0 \ $ and $ \ f_y \ = \ 4xy \ = \ 0 \ $ that there is a single critical point for the function at the origin. We see, however, that it sits at the junction of alternating regions of positive and negative function values, so $ \ (0,0) \ $ is a saddle point.

You presumably found the Lagrange equations

$$ 2y^2 \ - \ 4x \ \ = \ \ \lambda·2x \ \ \Rightarrow \ \ y^2 \ = \ (\lambda \ + \ 2)·x \ \ \ , \ \ \ 4xy \ \ = \ \ \lambda·2y \ \ \Rightarrow \ \ y·(2x \ - \ \lambda) \ \ = \ \ 0 \ \ . $$

It is best to deal with the second equation first, which produces the two cases $ \ y \ = \ 0 \ $ and $ \ \lambda \ = \ 2x \ \ : $

$ \mathbf{y = 0 : } \quad x^2 \ + \ 0^2 \ = \ 1 \ \ \Rightarrow \ \ x \ = \ \pm 1 \ \ \Rightarrow \ \ f(\pm 1 \ , \ 0) \ = \ 2·(\pm 1)·(0^2 - [\pm 1] \ ) \ = \ -2 \ \ , $

which are the absolute minima on the unit disk (for the condition $ \ x \ \ge \ 0 \ $ , we can discard the result for $ \ x = -1 \ $ ) ;

$ \mathbf{\lambda = 2x : } \quad y^2 \ = \ (2x \ + \ 2)·x \ = \ 2x^2 \ + \ 2x \ \ , $ which upon insertion into the constraint equation produces

$ x^2 \ + \ (2x^2 \ + \ 2x) \ = \ 3x^2 \ + \ 2x \ - \ 1 \ = \ (3x \ - \ 1)·(x \ + \ 1) \ = \ 0 \ \ . $ The root $ \ x \ = \ -1 \ $ tells us nothing new; the other root yields

$$ x \ = \ \frac13 \ \ \Rightarrow \ \ \left( \frac13 \right)^2 + \ y^2 \ = \ 1 \ \ \Rightarrow \ \ y^2 \ = \ \frac89 \ \ \Rightarrow \ \ y \ = \ \pm \frac{2 \sqrt{2}}{3} $$

$$ \Rightarrow \ \ f \left(\frac13 \ , \ \pm \frac{2 \sqrt{2}}{3} \right) \ = \ 2·\frac13 ·(\frac89 - \frac13 ) \ = \ 2·\frac13 ·\frac59 \ = \ \frac{10}{27} \ \ , $$

which are the absolute maxima. [The parabola $ \ y^2 \ = \ x \ $ intersects the unit circle at the points found from $ \ x^2 \ + \ x \ = \ 1 \ \Rightarrow \ x \ = \ \frac{-1 \ + \ \sqrt{1^2 \ - \ 4·1·(-1)}}{2} \ = \frac{-1 \ + \ \sqrt{5}}{2} \ \approx \ 0.618 \ , $ confirming that the maximal points we obtained are in the positive-value regions of $ \ f(x,y) \ \ . $ ]