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Wolfram says it's 800, but how to calculate it?

$$ \frac{25x}{x^2+1600x+640000} $$

  • 2
    Solve $\frac{d}{dx} f(x) = 0.$2012-11-03
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    @JenniferDylan - thank you, it works! but could you explain me why do you use derivative?2012-11-03
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    The derivative gives the slope of a function at each point. If you think of a hilltop, the ground is level there, so the derivative is zero. It is also zero at the bottom of a valley, so you need to check which you have. This is where the second derivative test comes from.2012-11-03
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    Maxima and minima occur at so-called [critical points](http://en.wikipedia.org/wiki/Maxima_and_minima). If $f(a)$ is a maximum value of $f(x)$, then you'd expect the values of $f(a-\epsilon)$, $f(a+\epsilon)$ to be smaller than $f(a)$. That is, you can draw a horizontal tangent at $f(a)$. A horizontal tangent has slope $= 0$, slope is 1st derivative.2012-11-03

4 Answers 4

1

Let $$\frac{25x}{(x^2+1600x+640000)}=y$$

or $$x^2y+x(1600y-25)+640000y=0$$

This is a quadratic equation in $x$.

As $x$ is real, the discriminant $(1600y-25)^2-4\cdot y\cdot 640000y\ge 0$

On simplification, $-128y+1\ge 0\implies 128y\le 1\implies y\le \frac1 {128}$

So, the maximum value of $y=\frac{25x}{(x^2+1600x+640000)}$ is $\frac1 {128}$

The value of $x$ for the maximum value of $y$ is $-\frac{1600y-25}{2y}$ where $y=\frac1 {128}$,

so $x$ will be $\frac{25-1600\cdot \frac 2{128} }{\frac2{128}}=\frac{25\cdot 128-1600}2=800$

Observe that $y$ does not have any lower limit\minimum value.

This approach can be applied to the expression like $\frac{ax^2+bx+c}{Ax^2+Bx+C}$

Reference: Minimum value of given expression

7

Note that $$ \frac{25x}{x^2+1600x+640 000}=\left(\frac{5}{\sqrt{x}+\frac{800}{\sqrt{x}}}\right)^2 $$ So maximum value is attained when denominator attains it minimal value. Now we use the following trick $$ \sqrt{x}+\frac{800}{\sqrt{x}}=\left(\sqrt[4]{x}-\frac{\sqrt{800}}{\sqrt[4]{x}}\right)^2+2\sqrt{800} $$ to see that minimal value of denominator attained when $$ \sqrt[4]{x}-\frac{\sqrt{800}}{\sqrt[4]{x}}=0 $$ i.e. when $x=800$.

5

It’s straightforward as a calculus problem. To solve it without calculus, note that

$$\frac{25x}{(x^2+1600x+640000)}=\frac{25x}{(x+800)^2}\;,\tag{1}$$

so the denominator is always positive, the the function has its maximum at some positive value of $x$. That maximum will occur where

$$\frac{(x+800)^2}{25x}=\frac1{25}\left(x+1600+\frac{640000}x\right)=64+\frac1{25}\left(x+\frac{640000}x\right)\tag{1}$$

has its minimum (over the range $x>0$). This in turn occurs where $x+\dfrac{640000}x$ has its minimum.

Now $x$ and $\frac{640000}x$ are a pair of numbers whose product is $640000=800^2$; if we set $x=800$, their sum is $1600$. Suppose that we set $x=800+a$ for some $a>0$; then

$$\begin{align*} x+\frac{640000}x&=800+a+\frac{640000}{800+a}\\ &=\frac{1280000+1600a+a^2}{800+a}\\ &=1600+\frac{a^2}{800+a}\\ &>1600\;. \end{align*}$$

Thus, $x=800$ gives us the minimum value of of $x+\frac{640000}x$, namely, $1600$, and hence the minimum value of $(2)$ and the maximum value of $(1)$. Substituting $x=800$ into $(1)$, we find that the maximum is $$\frac{25\cdot800}{1600^2}=\frac{25}{3200}=\frac1{128}\;.$$

2

By AM-GM inequality you have

$$\frac{x+800}{2} \geq \sqrt{800x}$$

Thus

$$\frac{1}{3200x} \geq \frac{1}{(x+800)^2}$$

Multiplying by $25x$ you get

$$\frac{25}{3200} \geq \frac{25}{(x+800)^2}$$

Equality is only when we have equality in AM-GM, that is when $x=800$.