multivariable-calculus
Find the maximum and minimum values of
$f(x,y)=xy-2x$ on the rectangle $-1\leq x \leq1 $ and $0\leq y \leq 1$.
I don't understand the approach. The solution manual suggests that the critical point is not inside my domain so maximum and minimum values of $f$ must be on one of the four boundary points. I don't understand how we get to this conclusion.
Obviously the function $f$ has no relative extreme inside the desired region $D$. In favt using the routine method we will find $(0,0)$ a saddle point in which $f_{xx}f_{yy}-(f_{xy}^2)<0$. Now consider $D$ and that $$x=\pm\sqrt{1-y^2}$$ Putting each parts $x=+$ and then $x=-$ separately we will find two one-variable functions $$f(y)=+y\sqrt{1-y^2}, ~~~f(y)=-y\sqrt{1-y^2}$$ I think you can find the relative extremes of these functions.... You have $4$ points as I plotted below:
You have the minimization optimization problem of the function $f(x,y) = xy$ over the space $S= \{ (x,y) \in \mathbb R^2 : x^2 + y^2 = 1\}$. A known way to deal with Lagrange multipliers is by the Kuhn-Tucker Lagrange method.
First of all, observe that $f(x,y)$ is continuous and smooth and that the space $S$ is compact. Thus, this means that there exists a minimum $(\bar{x},\bar{y})$ for $f(x,y)$ in $S$.
By the Kuhn-Tucker Lagrange method, we yield :
$$f_0(x,y) = xy, \; \; f_1(x,y)= x^2+y^2-1$$
and then the K.T.L. system :
$$\begin{cases} \nabla f_0 + \lambda_1\nabla f_1 = 0 \\ \lambda_1 f_1 =0 \end{cases} \Rightarrow \begin{cases} \begin{bmatrix} y \\ x \end{bmatrix} + \lambda_1\begin{bmatrix} 2x \\ 2y \end{bmatrix} =0 \\ \lambda_1(x^2 + y^2 -1) \;= 0\end{cases} $$
Check cases for $\lambda_1 = 0$ and $\lambda_1 >0$ and then you'll yield the same results. (Maximum is given for applying the same method for $-f(x,y)$ or simply you yield the same points as you did.
Now, if there existed another minimum or maximum, it should satisfy the K.T.L. problem. Since no other point satisfies it, these are all the minimums and maximums. Observing that you have two possible minimum and maximum points (since the values are equal) for $f(x,y)$ over $S$, this means that you have a total maximum and minimum at both of the points each time.
Best Answer
Check where the derivatives are zero in the domain given, if it's not zero anywhere in the domain then the maximum or the minimum would occur at the extreme values.
This will help http://personal.maths.surrey.ac.uk/st/S.Zelik/teach/calculus/max_min_2var.pdf