Answer:
The maximum value of f(x) occurs at:
![\displaystyle x = \frac{2a}{a+b}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20x%20%3D%20%5Cfrac%7B2a%7D%7Ba%2Bb%7D)
And is given by:
![\displaystyle f_{\text{max}}(x) = \left(\frac{2a}{a+b}\right)^a\left(\frac{2b}{a+b}\right)^b](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f_%7B%5Ctext%7Bmax%7D%7D%28x%29%20%3D%20%5Cleft%28%5Cfrac%7B2a%7D%7Ba%2Bb%7D%5Cright%29%5Ea%5Cleft%28%5Cfrac%7B2b%7D%7Ba%2Bb%7D%5Cright%29%5Eb)
Step-by-step explanation:
Answer:
Step-by-step explanation:
We are given the function:
![\displaystyle f(x) = x^a (2-x)^b \text{ where } a, b >0](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%28x%29%20%3D%20x%5Ea%20%282-x%29%5Eb%20%5Ctext%7B%20where%20%7D%20a%2C%20b%20%3E0)
And we want to find the maximum value of f(x) on the interval [0, 2].
First, let's evaluate the endpoints of the interval:
![\displaystyle f(0) = (0)^a(2-(0))^b = 0](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%280%29%20%3D%20%280%29%5Ea%282-%280%29%29%5Eb%20%3D%200)
And:
![\displaystyle f(2) = (2)^a(2-(2))^b = 0](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%282%29%20%3D%20%282%29%5Ea%282-%282%29%29%5Eb%20%3D%200)
Recall that extrema occurs at a function's critical points. The critical points of a function at the points where its derivative is either zero or undefined. Thus, find the derivative of the function:
![\displaystyle f'(x) = \frac{d}{dx} \left[ x^a\left(2-x\right)^b\right]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%27%28x%29%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%20%5Cleft%5B%20x%5Ea%5Cleft%282-x%5Cright%29%5Eb%5Cright%5D)
By the Product Rule:
![\displaystyle \begin{aligned} f'(x) &= \frac{d}{dx}\left[x^a\right] (2-x)^b + x^a\frac{d}{dx}\left[(2-x)^b\right]\\ \\ &=\left(ax^{a-1}\right)\left(2-x\right)^b + x^a\left(b(2-x)^{b-1}\cdot -1\right) \\ \\ &= x^a\left(2-x\right)^b \left[\frac{a}{x} - \frac{b}{2-x}\right] \end{aligned}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbegin%7Baligned%7D%20f%27%28x%29%20%26%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft%5Bx%5Ea%5Cright%5D%20%282-x%29%5Eb%20%2B%20x%5Ea%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft%5B%282-x%29%5Eb%5Cright%5D%5C%5C%20%5C%5C%20%26%3D%5Cleft%28ax%5E%7Ba-1%7D%5Cright%29%5Cleft%282-x%5Cright%29%5Eb%20%2B%20x%5Ea%5Cleft%28b%282-x%29%5E%7Bb-1%7D%5Ccdot%20-1%5Cright%29%20%5C%5C%20%5C%5C%20%26%3D%20x%5Ea%5Cleft%282-x%5Cright%29%5Eb%20%5Cleft%5B%5Cfrac%7Ba%7D%7Bx%7D%20-%20%5Cfrac%7Bb%7D%7B2-x%7D%5Cright%5D%20%5Cend%7Baligned%7D)
Set the derivative equal to zero and solve for <em>x: </em>
![\displaystyle 0= x^a\left(2-x\right)^b \left[\frac{a}{x} - \frac{b}{2-x}\right]](https://tex.z-dn.net/?f=%5Cdisplaystyle%200%3D%20x%5Ea%5Cleft%282-x%5Cright%29%5Eb%20%5Cleft%5B%5Cfrac%7Ba%7D%7Bx%7D%20-%20%5Cfrac%7Bb%7D%7B2-x%7D%5Cright%5D)
By the Zero Product Property:
![\displaystyle x^a (2-x)^b = 0\text{ or } \frac{a}{x} - \frac{b}{2-x} = 0](https://tex.z-dn.net/?f=%5Cdisplaystyle%20x%5Ea%20%282-x%29%5Eb%20%3D%200%5Ctext%7B%20or%20%7D%20%5Cfrac%7Ba%7D%7Bx%7D%20-%20%5Cfrac%7Bb%7D%7B2-x%7D%20%3D%200)
The solutions to the first equation are <em>x</em> = 0 and <em>x</em> = 2.
First, for the second equation, note that it is undefined when <em>x</em> = 0 and <em>x</em> = 2.
To solve for <em>x</em>, we can multiply both sides by the denominators.
![\displaystyle\left( \frac{a}{x} - \frac{b}{2-x} \right)\left((x(2-x)\right) = 0(x(2-x))](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cleft%28%20%5Cfrac%7Ba%7D%7Bx%7D%20-%20%5Cfrac%7Bb%7D%7B2-x%7D%20%5Cright%29%5Cleft%28%28x%282-x%29%5Cright%29%20%3D%200%28x%282-x%29%29)
Simplify:
![\displaystyle a(2-x) - b(x) = 0](https://tex.z-dn.net/?f=%5Cdisplaystyle%20a%282-x%29%20-%20b%28x%29%20%3D%200)
And solve for <em>x: </em>
![\displaystyle \begin{aligned} 2a-ax-bx &= 0 \\ 2a &= ax+bx \\ 2a&= x(a+b) \\ \frac{2a}{a+b} &= x \end{aligned}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbegin%7Baligned%7D%202a-ax-bx%20%26%3D%200%20%5C%5C%202a%20%26%3D%20ax%2Bbx%20%5C%5C%202a%26%3D%20x%28a%2Bb%29%20%5C%5C%20%20%5Cfrac%7B2a%7D%7Ba%2Bb%7D%20%26%3D%20x%20%20%5Cend%7Baligned%7D)
So, our critical points are:
![\displaystyle x = 0 , 2 , \text{ and } \frac{2a}{a+b}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20x%20%3D%200%20%2C%202%20%2C%20%5Ctext%7B%20and%20%7D%20%5Cfrac%7B2a%7D%7Ba%2Bb%7D)
We already know that f(0) = f(2) = 0.
For the third point, we can see that:
![\displaystyle f\left(\frac{2a}{a+b}\right) = \left(\frac{2a}{a+b}\right)^a\left(2- \frac{2a}{a+b}\right)^b](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%5Cleft%28%5Cfrac%7B2a%7D%7Ba%2Bb%7D%5Cright%29%20%3D%20%5Cleft%28%5Cfrac%7B2a%7D%7Ba%2Bb%7D%5Cright%29%5Ea%5Cleft%282-%20%5Cfrac%7B2a%7D%7Ba%2Bb%7D%5Cright%29%5Eb)
This can be simplified to:
![\displaystyle f\left(\frac{2a}{a+b}\right) = \left(\frac{2a}{a+b}\right)^a\left(\frac{2b}{a+b}\right)^b](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%5Cleft%28%5Cfrac%7B2a%7D%7Ba%2Bb%7D%5Cright%29%20%3D%20%5Cleft%28%5Cfrac%7B2a%7D%7Ba%2Bb%7D%5Cright%29%5Ea%5Cleft%28%5Cfrac%7B2b%7D%7Ba%2Bb%7D%5Cright%29%5Eb)
Since <em>a</em> and <em>b</em> > 0, both factors must be positive. Thus, f(2a / (a + b)) > 0. So, this must be the maximum value.
To confirm that this is indeed a maximum, we can select values to test. Let <em>a</em> = 2 and <em>b</em> = 3. Then:
![\displaystyle f'(x) = x^2(2-x)^3\left(\frac{2}{x} - \frac{3}{2-x}\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%27%28x%29%20%3D%20x%5E2%282-x%29%5E3%5Cleft%28%5Cfrac%7B2%7D%7Bx%7D%20-%20%5Cfrac%7B3%7D%7B2-x%7D%5Cright%29)
The critical point will be at:
![\displaystyle x= \frac{2(2)}{(2)+(3)} = \frac{4}{5}=0.8](https://tex.z-dn.net/?f=%5Cdisplaystyle%20x%3D%20%5Cfrac%7B2%282%29%7D%7B%282%29%2B%283%29%7D%20%3D%20%5Cfrac%7B4%7D%7B5%7D%3D0.8)
Testing <em>x</em> = 0.5 and <em>x</em> = 1 yields that:
![\displaystyle f'(0.5) >0\text{ and } f'(1)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%27%280.5%29%20%3E0%5Ctext%7B%20and%20%7D%20f%27%281%29%20%20%3C0)
Since the derivative is positive and then negative, we can conclude that the point is indeed a maximum.
Therefore, the maximum value of f(x) occurs at:
![\displaystyle x = \frac{2a}{a+b}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20x%20%3D%20%5Cfrac%7B2a%7D%7Ba%2Bb%7D)
And is given by:
![\displaystyle f_{\text{max}}(x) = \left(\frac{2a}{a+b}\right)^a\left(\frac{2b}{a+b}\right)^b](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f_%7B%5Ctext%7Bmax%7D%7D%28x%29%20%3D%20%5Cleft%28%5Cfrac%7B2a%7D%7Ba%2Bb%7D%5Cright%29%5Ea%5Cleft%28%5Cfrac%7B2b%7D%7Ba%2Bb%7D%5Cright%29%5Eb)