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I want to find the min of this function $f(x,y)= x^2-x+y^2$ graphically under the constraint $x \geq 2$ and $x+y \geq 1$ i find the area of the constraints.

I know how to find the solution analytically but i want to find it graphically. I notice $x^2-x+y^2$ can be the equation of the cercle of center (0.5, 0) and radius 0.5 but after this i don't know what to do, what does it mean to minimise a circle on a region ?

Farouk Deutsch
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If you put it equal to zero, then sure. That is called a level set of the function. The key here is imposing it equal to, say, $\lambda$ and understand what are the max and min values of the parameter for which your level set has a non empty intersection with the domain you are working with. In this case, since you can reduce to the sum of two squares, it is really easy to guess the minimum. And the sup is also quite straightforward, can you see them?

HINT: Whatever the $\lambda$, the center of the circle is bound to be the one you’ve found. So, if increasing/decreasing $\lambda$ means increasing/decreasing the radius...

tommy1996q
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  • i have this $(x-1/2)^2 + y^2 -1/4= \lambda$ and so we want to minimise lambda ? – Farouk Deutsch May 28 '20 at 22:16
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    $f(x,y)= \lambda$ gives you $(x -1/2)^2 +y^2= \lambda + 1/4$. You gotta find the minimum radius (which as you can see depends on $\lambda$) such that that circle “touches” you domain. I am not saying tangent because the domain seems to have a “spike”, and those points don’t have a tangent. – tommy1996q May 28 '20 at 22:41
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It can be shown that $f(x,y) = (x-\frac 1 2)^2 + y^2 - \frac 1 4$.

Consider the curve defined by the equation $(x-\frac 1 2)^2 + y^2 - \frac{1}{4} = \lambda$ with $\lambda \geq -1/4.$ This is the equation of a circle with radius $\sqrt{\lambda + \frac 1 4}$.

If you draw this circle then all the points $(x,y)$ inside the circle will be such that $(x-\frac 1 2)^2 + y^2 < \frac 1 4 + \lambda$ i.e. the points $(x,y)$ such that $f(x,y)<\lambda.$

The constraints of your problem are $ x \geq 2$ and $x + y \geq 1$. The $(x,y)$ that satisfy these two conditions are exactly the points that are to the right of the line $x = 2$ and above the line $y = 1 - x$. Color this region in blue.

If there is a point $(x',y')$ that is both inside the circle and in the blue zone then $f(x,y) < \lambda$ therefore $f$ can take smaller value than $\lambda$ under the constraints.

You're looking for a circle which touches the blue area but is such that are no points inside the the circle that are in the blue area. This circle is the circle with radius $3/2$ (that touches the line $x = 2$.)

This means $f$ takes its minimal value under the constraints at $(x,y) = (2,0)$ and this minimal value is $2.$

Digitallis
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