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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!

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Is taking the derivative allowed? –  Ross Millikan Aug 2 '11 at 15:49
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You could write $u=x^2$, $x=\sqrt u$ to get $(10-\sqrt u)(10-\sqrt {81-u})$, which simplifies the derivative somewhat, but then you have to treat $x=0$ separately. You could then use the symmetry by writing $u=81/2+s$ to get $(10-\sqrt{81/2+s})(10-\sqrt{81/2-s})$, which allows you to guess the maximum at $s=0$, corresponding to $x=\sqrt{81/2}$, but I don't see how you could get the two minima without actually differentiating the function. By the way, assuming we're talking about a real-valued function, this function is ill-defined for $x>9$. –  joriki Aug 2 '11 at 16:36
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How do we talk about the maximum or minimum of a function when it can have values in both the reals and the complex numbers? –  Doug Spoonwood Aug 2 '11 at 16:37
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@joriki: I don't even see a reasonable calculus approach, at least using hand computations (which I've put some effort into), although assuming I typed things correctly, WolframAlpha says all extrema are quadratic irrationals. Maybe someone can do something with the following observation. Consider the circle $x^2 + y^2 = 81$ inside the square with sides $x = \pm 10$ and $y = \pm 10$. The question asks for the extrema of the product of a point's distances from the upper side and the right side of the square as the point traces through 1st quadrant portion of the circle. –  Dave L. Renfro Aug 2 '11 at 17:44
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@Dave: Starting from the form $(10-\sqrt{81/2+s})(10-\sqrt{81/2-s})$, you can take the derivative, multiply through with $2\sqrt{81/2+s}\sqrt{81/2-s}$, multiply out, take the two remaining square roots to one side, square, take the remaining root to one side and square again. The result is a biquadratic equation for $s$ with the solutions $s=0$ and $s=\pm5\sqrt{62}$, so the minima are at $x=\sqrt{81/2\pm5\sqrt{62}}$, which W|A simplifies to $5\pm\sqrt{31/2}$. Not a well-chosen question for high-school students if you ask me... :-) –  joriki Aug 2 '11 at 18:12
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5 Answers

up vote 5 down vote accepted

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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Good point! This looks much easier. –  Julie Aug 4 '11 at 14:45
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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 $x = 9\sin(\phi+45^\circ)=(9/\sqrt{2})(\cos\phi\pm\sqrt{1-\cos^2\phi})$.
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Thanks, probably I should try to use trigonometric function –  Julie Aug 3 '11 at 1:17
    
Nice solution. This is actually slightly suggested by how the function is written ($81$ as $9^2$), so perhaps this is how they intended it to be solved. –  joriki Aug 3 '11 at 4:29
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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:

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The question got categorized as an algebra question and mentioned "high school", so are derivatives allowed for an answer? –  Doug Spoonwood Aug 2 '11 at 16:33
    
@Doug: Well, other users can see other ways of doing it. WHile looking at this question, I didn't really see the tag. –  user9413 Aug 2 '11 at 16:37
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In any case, you don't need the second derivative. On a closed interval, the max/min will occur at endpoints or critical points. –  The Chaz 2.0 Aug 2 '11 at 17:42
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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.

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I agree with the graphical interpretation. The maximum and minimum corresponds to the maximum/minimum area of the rectangle –  Julie Aug 3 '11 at 1:16
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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

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The derivative is wrong, and some of the rest is hard to follow. It's not apparent what you're deducing from what (i.e. what the "if"s refer to) and what your variable names mean (x1, x2, x3). Also, there are roots of any multiplicitly, not just single, double and triple ones. –  joriki Aug 2 '11 at 17:43
    
i‘ve corrected the derivation Xe is the variable name for the extrema (Min/Max). @jorki yes there are three roots but for maxima and minima of an curve you take onle the non complex parts for example $\sqrt{-9}$ is a complex result 3i (i is an imaginary number (don't know the right english word)) –  RenHoek Aug 2 '11 at 19:49
    
Sorry for everyone. I should narrow x to [0,9]. –  Julie Aug 3 '11 at 1:13
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