Find the maximum and minimum values in a range

Question

I'm trying to understand how to find the minimum & maximum values of this function:

$$ f(x,y) = xy-y^2 $$

In the following range D:

$$ D = \{(x,y) \in R^2 : 0 \leq x \leq 1, |y| \leq x^2 \} $$

Obviously I tried to use Lagranage multipliers, but I was a little confused about the absolute value. Should I divide it to two different equastion systems, one for positive y and one for negative?

Thanks in advance.

Answer 1

observe that $$f_x=y$$ and $$f_y=x-2y$$ thus we get the solution $$x=0,y=0$$ from the system $$f_x=0$$ and $$f_y=0$$

Answer 2

The region specification $ \ |y| \leq x^2 \ $ is a "shorthand": it could as easily have been written as $ \ -x^2 \ \leq \ y \ \leq \ x^2 \ $ . So the region lies between two parabolas and between the origin and the vertical line $ \ x = 1 \ \ , $ as seen in the graph below.

For the interior region, you would use Lagrange multipliers as you proposed; the result shown by Dr. Sonnhard Graubner tells us that

$$ f_x \ = \ y \ = \ 0 \ \ , \ \ f_y \ = \ x - 2y \ = \ 0 \ \ \Rightarrow \ \ x \ = \ 2y \ \ . $$

So there is a single critical point in the interior of the region at $ \ (0,0) \ . $ Since the function is $ \ f(x,y) \ = \ xy \ - \ y^2 \ = \ y · (x-y) \ , $ we have $ \ \ f(x,0) \ = \ 0 \ $ and $ \ \ f(x,x) \ = \ 0 \ $ . So this critical point is a little peculiar in that it lies at the intersection of two lines (the $ \ x-$ axis and $ \ y \ = \ x $ ) for which the function is zero. The index $ \ f_{xx}f_{yy} \ - \ (f_{xy})^2 \ = \ 0 · (-2) \ - \ 1^2 \ = \ -1 \ < \ 0 \ $ is constant for the function, suggesting that $ \ (0,0) \ $ is a saddle point. [Indeed, it is at the junction of four regions of alternating sign as one circulates around the origin.] Since for our region, $ \ x \ > \ y \ $ , $ \ f(x,y) \ > \ 0 \ $ for $ \ y \ > \ 0 \ $ and $ \ f(x,y) \ < \ 0 \ $ for $ \ y \ < \ 0 \ . $

It is when we check the boundary of the region for extremal points that we need to work with the two individual parabolas (and the vertical line).

On $ \ x \ = \ 1 \ $ , the function becomes $ \ \ f(1,y) \ = \ 1·y \ - \ y^2 \ , $ so we have $$ \ \frac{df}{dy} |_{x=1} \ = \ 1 - 2y \ = \ 0 \ \ \Rightarrow \ \ y \ = \ \frac{1}{2} \ \ , $$

giving us an extremal point (local maximum) at $ \ \left(1 \ , \ \frac{1}{2} \right) \ $ with $ \ \ f \left(1 \ , \ \frac{1}{2} \right) \ = \ \frac{1}{2} \ - \ \left( \frac{1}{2} \right)^2 \ = \ \frac{1}{4} \ \ . $ [This is where we might have expected to find such a point from our earlier result of $ \ f_y \ = \ 0 \ $ on the line $ \ x \ = \ 2y \ . $ ]

Along the "upper" parabolic boundary, $$ \ \ f(x,x^2) \ = \ x·x^2 \ - \ (x^2)^2 \ \ = \ \ x^3 \ - \ x^4 \ \ \Rightarrow \ \ \frac{df}{dy} |_{x^2} \ = \ x^2 · (3 - 4x) \ = \ 0 \ \ , $$

so there is a local maximum at $ \ \left(\frac{3}{4} \ , \ \frac{9}{16} \right) \ $ with $ \ \ f \left(\frac{3}{4} \ , \ \frac{9}{16} \right) \ = \ \frac{3}{4} · \frac{9}{16} \ - \ \left( \frac{9}{16} \right)^2 \ = \ \frac{27}{256} \ \approx \ 0.105 \ \ . $

For the "lower" parabolic boundary, $$ \ \ f(x,-x^2) \ = \ x·(-x^2) \ - \ (-x^2)^2 \ \ = \ \ -x^3 \ - \ x^4 \ \ \Rightarrow \ \ \frac{df}{dy} |_{-x^2} \ = \ -x^2 · (3 + 4x) \ = \ 0 \ \ , $$

for which the solution is outside of our region.

It remains to test the "vertex" $ \ (1 , -1) \ $ , as we already know that $ \ f (0 , 0) \ = \ f(1 , 1) \ = \ 0 \ $ :

$ \ \ f (1 , -1 ) \ = \ 1 · (-1) \ - \ ( -1 )^2 \ = \ -2 \ \ . $

The absolute maximum in the region is thus $ \ \ f \left(1 \ , \ \frac{1}{2} \right) \ = \ \frac{1}{4} \ \ $ and the absolute minimum is $ \ \ f (1 , -1 ) \ = \ -2 \ \ . $