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I have an assignment I will hand in for grading.

First part

I want to see if I have understood the consept correctly.

Determine max and min of $f(x,y)=xy $ on the triangle T with the corners $(0, 0), (0, 1)\:and\:(1, 0)$

I derived and got a critical point at $(0,0)$ which I evaluated to $0$.

Secondly I inspected the boundary lines of which only the hypotenuse was interesting x+y=1 where I only found one critical point and that was larger than 0.

Is it correct minimum value was 0?

Second part

Are there any analytical online tools where I can write functions and boundaries and get the max and min values?

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For the first part, you should also check the vertices of the triangle, for the same reason you check the boundaries of the domain. –  Daryl Sep 27 '12 at 23:35

1 Answer 1

I was surprised to find this question abandoned, so I thought I'd finish it.

It is clear that Farmor has already noted that setting $ \ \frac{\partial f}{\partial x} = y \ $ and $ \ \frac{\partial f}{\partial y} = x \ $ equal to zero locates the critical point at $ \ (0,0) \ , $ at which $ \ f(0,0) = 0 \ . $ The function is also zero at the boundaries of the triangle and at all three vertices (thus responding to Daryl's remark, although OP may well have already done that if that is what was meant by having inspected the boundaries). So the minimum value of the function in the region is zero, but that is not localized to a "critical point".

As for the boundary on $ \ x + y = 1 \ $ , the single-variable approach is to write $ \ f(x,y) = xy \ $ as $ \ f(x) = x \cdot (1-x) = x - x^2 \ , $ which has its maximum at $ \ f(\frac{1}{2}) = \frac{1}{4} \ $ (so identified since $ \ f'(\frac{1}{2}) = -2 \cdot \frac{1}{2} < 0 \ $ ) ; this can be found even without calculus by noting the properties of the "downward-facing" parabola $ \ y = x - x^2 \ . $

Alternatively, we can use Lagrange multipliers with the function $ \ f(x,y) = xy \ $ under the constraint $ \ g(x,y) = x + y - 1 \ , $ producing the equations

$$\begin{array}{cc}f_x \ = \ \lambda \ \cdot \ g_x\\f_y \ = \ \lambda \ \cdot \ g_y\end{array} \ \Rightarrow \ \begin{array}{cc}x \ = \ \lambda \ \cdot \ 1\\y \ = \ \lambda \ \cdot \ 1\end{array} $$

$$ \Rightarrow \ \lambda \ = \ x \ = \ y \ = \ \frac{1}{2} \ . $$

So we find what proves to be a local and absolute maximum, $ \ f(\frac{1}{2} , \frac{1}{2}) = \frac{1}{4} . $ The standard discriminant $ \ f_{xx} \cdot f_{yy} - ( f_{xy} )^2 = 0 \cdot 0 - 1^2 \ $ is of no help, however, in characterizing critical points for this function.

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