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I've been asked to find the minimum and maximum of the following function:

$f(x,y) = x^2+y^2-x+1/4$

On the region or restriction defined as:

$D$={${(x,y)\in\mathbb{R}^2:x^2+y^2\leq1; x+y\leq0}$}

First, I observed that $f$ is continuos, and after I did the graph of the region where I will study the function, to check geometrically that the intersection of the two conditions I have defined, is closed and bounded, factor which assure that exist extrems because of the extreme value theorem.

Then, I studied the interior of the region with $\nabla f(x,y)=(0,0)$ and I get the critical point $(1/2,0)$

And now I'm having problems to could study the borders of the region. I parametrized one of them through the following curve: $\sigma (t)=(t,-t)$ and then, I compose $f$ with it, $f$$\circ$$\sigma(t)$$= 2t^2-t+1/4$. Once I did this, I get from the ($f$$\circ$$\sigma)'(t)=0$ the critical point $(1/4,-1/4)$. But, I don't know how to parametrize the rest, that is the border of a half part of the unitary circle. Any idea?

Neisy Sofía Vadori
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3 Answers3

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A much simpler approach is to note that $f(x,y)=(x-\frac 12)^2+y^2$, the squared distance from $(\frac 12,0)$. Your critical point is the global minimum, but outside the region of interest. The minimum in your region will be the point closest to $(\frac 12,0)$. You can find that by taking the perpendicular to $x+y=0$ that goes through your point. That is $y=x+\frac 12$, so the point is $(\frac 14,-\frac 14)$. For the maximum you want the farthest point from $(\frac 12,0)$ which would be $(-1,0)$.

To do what you were trying, you can parameterize the circle with $x=\cos t, y=\sin t$ with $\frac {3 \pi}4 \le t \le \frac {7 \pi}4$

Ross Millikan
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  • Great! Thanks! And what about the the curve $(\sqrt{ 2 }$$/2,-\sqrt{ 2 }$$/2)$ deduced from where $t$ is restricted, isn't also a critical point? – Neisy Sofía Vadori Jun 15 '17 at 15:38
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    It is a place you should check as it is on the boundary. It represents $t=\frac {7\pi}4$ in my parameterization of the circle. It will not be a local maximum or minimum, as you can reduce the function by going up the $y=-x$ line and increase it by going left. – Ross Millikan Jun 15 '17 at 15:52
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As Ross Millikan points out, you don not need calculus.

But since you are taking multivariate calculus, and you want to use your new tools.

You have found that there is no local minimum in the interior of the region.

You have found the point $(\frac 14,\frac 14)$ on the line $x+y = 0$

The rest of the boarder is a section of a circle. How do you paramaterize a circle?

There is more that one way to do it, but simplest is usually $(\cos t, \sin t)$

$f(t) = \frac 54 - \cos t$

Which will be minimized when $\cos t$ is maximized, and maximized when t is minimized.

Restrict $t$ to $[\frac {3\pi}{4},\frac {7\pi}{4}]$ to comply with $x+y\le 0$

$(-1,0),(\frac {\sqrt2}{2}, -\frac {\sqrt2}{2})$ are your points to check.

And you should always check the corners. Check $(-\frac {\sqrt2}{2}, \frac {\sqrt2}{2})$

Doug M
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If the objective function is rearranged, the contour set will be a set of circles with radius $\sqrt{z-\frac12}$: $$\left(x-\frac12\right)^2+y^2=z-\frac12$$

The objective is to find the maximum and minimum of $z$, that is the largest and smallest contour circle.

The feasible region and the extreme contour sets (circles) are shown below: enter image description here

Thus, the max $z(-1,0)=\frac94$ amd the min $z(\frac14,-\frac14)=\frac18$.

farruhota
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