<s>[INST] Local minimum and maximum of the function Can anyone help me to solve the following question?
maximize and minimize the function $(10-x)(10-\sqrt{9^2-x^2})$ over $x\in[0,10]$
This is a high school question, so is there any simple trick help solve it? Thanks!
 [/INST] We give a short argument for both maximum and minimum.   
To bring out the symmetry, let $y=\sqrt{81-x^2}$.
We study the behaviour of $(10-x)(10-y)$, that is, of
$$xy-10(x+y)+100.\qquad\qquad\text{(Expression $1$)}$$
We will find the maximum and minimum values of Expression $1$, given that
$x^2+y^2=81$ and $x \ge 0$, $y \ge 0$.
Let $w=x+y$ and $xy=v$.
We are interested in the behaviour of
$$v-10w+100. \qquad\qquad\text{(Expression $2$)}$$
But since $(x+y)^2-2xy=x^2+y^2$, we have $w^2-2v=81$.
So we are interested in the behaviour of
$$\frac{w^2-81}{2} -10w +100, \quad\text{that is, of}$$
$$\frac{1}{2}\left(w^2-20w+119\right).  $$
Complete the square.  We get
$$\frac{1}{2}\left((w-10)^2+19\right).\qquad\qquad\text{(Expression $3$)}$$
Now it's over. There is a local minimum at $w=10$. 
The minimum value is $19/2$.
The maximum is reached where $w$, that is, $x+y$, reaches a maximum subject to $x^2+y^2=81$.  Since $(x+y)^2+(x-y)^2=2(x^2+y^2)=162$, the maximum value of $(x+y)^2$, and hence of $x+y$, occurs where $x=y$. 
We can if we wish find the values of $x$ at which the minimum is reached.  We need to solve the system $x+y=10$, $x^2+y^2=81$. That gives $2xy=10^2-81=19$, so $(x-y)^2=81-19=62$, and therefore $x-y =\pm \sqrt{62}$, and now we can find $x$ and $y$. 
Note on Symmetry:  All the way through, we have preserved symmetry between $x$ and $y$. We introduced symmetry in the initial setup, and every step involved only symmetric functions of $x$ and $y$.  Symmetry allowed the easy identification of $x+y$ as a key parameter.   
The formal algebraic symmetry comes, in this case, from the underlying geometry. For the problem posed by the OP is fundamentally geometric. It has to do with the interaction between a circle and a rectangular hyperbola.  
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