Mathematics, 30.07.2021 08:30 cadeduncan71

If a and b are positive numbers, find the maximum value of f(x) = x^a(2 − x)^b on the interval 0 ≤ x ≤ 2.

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Answer from: colochaortiz20p7cajw

The maximum value of f(x) occurs at:

$\displaystyle x = \frac{2a}{a+b}$

And is given by:

$\displaystyle f_{\text{max}}(x) = \left(\frac{2a}{a+b}\right)^a\left(\frac{2b}{a+b}\right)^b$

Step-by-step explanation:

We are given the function:

$\displaystyle f(x) = x^a (2-x)^b \text{ where } a, b 0$

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:

$\displaystyle f(0) = (0)^a(2-(0))^b = 0$

And:

$\displaystyle f(2) = (2)^a(2-(2))^b = 0$

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:

$\displaystyle f'(x) = \frac{d}{dx} \left[ x^a\left(2-x\right)^b\right]$

By the Product Rule:

$\displaystyle \begin{aligned} f'(x) &= \frac{d}{dx}\left[x^a\right] (2-x)^b + x^a\frac{d}{dx}\left[(2-x)^b\right]\\ \\ &=\left(ax^{a-1}\right)\left(2-x\right)^b + x^a\left(b(2-x)^{b-1}\cdot -1\right) \\ \\ &= x^a\left(2-x\right)^b \left[\frac{a}{x} - \frac{b}{2-x}\right] \end{aligned}$

Set the derivative equal to zero and solve for x:

$\displaystyle 0= x^a\left(2-x\right)^b \left[\frac{a}{x} - \frac{b}{2-x}\right]$

By the Zero Product Property:

$\displaystyle x^a (2-x)^b = 0\text{ or } \frac{a}{x} - \frac{b}{2-x} = 0$

The solutions to the first equation are x = 0 and x = 2.

First, for the second equation, note that it is undefined when x = 0 and x = 2.

To solve for x, we can multiply both sides by the denominators.

$\displaystyle\left( \frac{a}{x} - \frac{b}{2-x} \right)\left((x(2-x)\right) = 0(x(2-x))$

Simplify:

$\displaystyle a(2-x) - b(x) = 0$

And solve for x:

$\displaystyle \begin{aligned} 2a-ax-bx &= 0 \\ 2a &= ax+bx \\ 2a&= x(a+b) \\ \frac{2a}{a+b} &= x \end{aligned}$

So, our critical points are:

$\displaystyle x = 0 , 2 , \text{ and } \frac{2a}{a+b}$

We already know that f(0) = f(2) = 0.

For the third point, we can see that:

$\displaystyle f\left(\frac{2a}{a+b}\right) = \left(\frac{2a}{a+b}\right)^a\left(2- \frac{2a}{a+b}\right)^b$

This can be simplified to:

$\displaystyle f\left(\frac{2a}{a+b}\right) = \left(\frac{2a}{a+b}\right)^a\left(\frac{2b}{a+b}\right)^b$

Since a and b > 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 a = 2 and b = 3. Then:

$\displaystyle f'(x) = x^2(2-x)^3\left(\frac{2}{x} - \frac{3}{2-x}\right)$

The critical point will be at:

$\displaystyle x= \frac{2(2)}{(2)+(3)} = \frac{4}{5}=0.8$

Testing x = 0.5 and x = 1 yields that:

$\displaystyle f'(0.5) 0\text{ and } f'(1)$

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:

$\displaystyle x = \frac{2a}{a+b}$

And is given by:

$\displaystyle f_{\text{max}}(x) = \left(\frac{2a}{a+b}\right)^a\left(\frac{2b}{a+b}\right)^b$

Answer from: Quest

answer: c, the green line

step-by-step explanation:

there is no black line, eliminate b.

simple interest implies periodic payments, but the red line shows no payments, eliminate d.

simple interest has equal payments. if paid say every day, the graph is about a straight line (actually a stairstep).

the yellow line is not straight, eliminate a.

check: the green line is straight, answer is c.

simple interest with positive interest rate gives a straight line sloping up to the right. green line:

g(t) = $100 + ($200-$100) ÷ (20 years) × (t years)

= $100 + (4%/year) × ($100)(t years)

incidentally, the yellow line curving upward represents continuously compounded interest

y = $100 × 2^(t years / (14 assuming the yellow line hits $200 at 14 years. 2^0 is 1, so y(0) = $100; 2^(14/14) is 2, so y(14) = $200.

The graph below shows the value of a $100 deposited into three different accounts over a period of 2

Answer from: Quest

A. but. i don’t nohxbzbz

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If a and b are positive numbers, find the maximum value of f(x) = x^a(2 − x)^b on the interval 0 ≤ x...

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