4

I know it's a bit short ,

But I have no idea at all how to solve such question ..

I'd be very glad if you can give me some help HOW to solve it.

Given fuction $f(x, y) = x - x^2 - xy^2$, and I need to find it's minimum and maximum value in the domain $D = \{(x, y) : x^2 + y^2 \le 4, x \ge 0\}$.

How can I solve it?

Is that somehow related with Lagrange multipliers?

Billie
  • 3,449

2 Answers2

6

You do not have to use Lagrange Multipliers, although you should repeat this exercise in order to duplicate the answer using that approach.

We are given:

$$f(x, y) = x - x^2 - xy^2, ~~D = \{(x, y) : x^2 + y^2 \le 4, x \ge 0\}$$

We need to find all of the candidate Critical Points (CPs).

  • Interior CPs:

$$f_x = 1-2x -y^2 = 0, f_y = -2xy = 0 \rightarrow (x,y) = \left(\dfrac{1}{2},0\right), (0, 1), (0, -1)$$

  • Boundary CPs:

$$D = \{(x, y) : x^2 + y^2 \le 4\} \rightarrow y \le \pm~\sqrt{4-x^2}~~\mbox{with endpoints}~~(x,y) = (0, \pm~ 2)$$

  • $f(x,y)$ becomes $$f = x-x^2-xy^2 = x-x^2-x(4-x^2) = 0 \rightarrow x = \dfrac{1}{3}(1 \pm ~ 10)$$

We reject the negative $x$ point because of the constraint that $x \ge 0$, so $y = \pm~ \sqrt{4-x^2} = \pm ~ 1.4405$.

Thus, we have the seven critical points to test by plugging them in $f(x, y)$. For the four points where $x = 0$, the function value equals zero, so we only need to test the other three points.

We find:

  • Global Maximum at $(x, y) = \left(\dfrac{1}{2},0\right)$, that is $\dfrac{1}{4}$.
  • Global Minimum at $(x, y) = (1.38743, -1.4405)$ and $(1.38743, 1.4405)$, that is $-3.4165$.
Amzoti
  • 56,093
2

In the interior of the semicircle, use the usual test, namely, you want $Df(x,y)=0$. In the boundary, you need to maximize or minimize a one variable function: either set $x=0$ or $x^2+y^2=4$ and substitute.

Pedro
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