Local minimum and maximum of the function

Question

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!

Answer 1

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.

Answer 2

I think a "precalculus"-typed solution is bound to be tedious, but you may try the following approach.

1. Note that the function is non-real on the interval (9,10], so the function domain should be first corrected to [0,9].
2. Put $x=9\sin\theta$ and transform the objective function to $(10-9\sin\theta)(10-9\cos\theta)$ with $0\le\theta\le 90^\circ$.
3. So, up to a multiple, the objective function is of the form $(t-\sin\theta)(t-\cos\theta)$, where $t=\frac{10}{9}$ and $0\le\theta\le 90^\circ$.
4. Put $\theta= 45^\circ+\phi$. Then, up to a suitable multiple, the objective function takes the form of $(u-\cos\phi-\sin\phi)(u-\cos\phi+\sin\phi)$, where $u=\frac{10\sqrt{2}}{9}$ and $-45^\circ\le\phi\le 45^\circ$.
5. The function is equal to $(u-\cos\phi)^2-\sin^2\phi = 2\cos^2\phi-2u\cos\phi+(u^2-1)$, which is quadratic in $\cos\phi$. Now you can try to find its maximum and minimum with $\cos\phi\in[\frac{1}{\sqrt{2}}, 1]$.
6. Edit: Having $\cos\phi$, we may compute $x = 9\sin(\phi+45^\circ)=(9/\sqrt{2})(\cos\phi\pm\sqrt{1-\cos^2\phi})$.

Answer 3

Let $f(x)=(10-x)\cdot (10-\sqrt{81-x^{2}})$. First solve for $f'(x) =0$. Then find $f''(x)$. If $f''(x) < 0$, then you have a maximum and if $f''(x) > 0$, then you have a minimum.

Please see:

• http://en.wikipedia.org/wiki/Second_derivative_test

• http://en.wikipedia.org/wiki/First_derivative_test

Answer 4

Perhaps this is a graphing question. Some high school teachers allow the use of calculators to get numerical approximations for certain problems. I can't see how else this problem is possible without calculus.

Answer 5

Generally, one can note the following:

There are single, double and triple roots. For a simple root of the graph intersects the x-axis Example x1=0 With a double root touch it and indicating an extreme point. Example x1=0,x2=0 Tripple a root point is a saddle point Example x1=0,x2=0,x3=0.

if your equatiion:

$f(x)=(10-x)\cdot (10-\sqrt{81-x^{2}})$

$f(x)=0$ $(10-x)=0 $

if $ x=10$

$(10-\sqrt{81-x^{2}})=0$

$10 = \sqrt{81-x^{2}}$ | ^2

$100 = 81-x^{2}$

$-19 = x^{2}$

$\sqrt{-19}=x$ ERROR ! you will get complex roots... not useable for Maxima and Minima

root = 10

derivation of $f(x) = y$ is

$y' = - \frac{(-10 + x) x)}{ \sqrt{81 - x^2}} + \sqrt{81 - x^2} -10$

To get the extrema you had to $y'=0$ Xe = root(s) of $y'=0$

and put the roots (Xe) if any there in $y''(Xe)$ $y''(Xe) < 0 $ it's a Maximum $y''(Xe) > 0 $ it's a Minimum
$y''(Xe) = 0 $ Xe is not an extrema