What Value of x Maximizes the Function y=axln(b/x)?

[Karol]

Homework Statement

$$y=a\cdot x\cdot ln\left(\frac{b}{x}\right)$$
The derivative should be 0 (to maximize), what's x?

Homework Equations

$$(ln\:x)'=\frac{1}{x}$$
$$(x^a)'=ax^{(a-1)}$$
$$(uv)'=u'v+v'u$$

The Attempt at a Solution

$$\dot y=a \left[ ln \left( \frac{b}{x} \right)-x\frac{x}{b}x^{-2} \right]=a \left[ ln \left( \frac{b}{x} \right)-\frac{1}{b} \right]$$
$$\dot y=0 \rightarrow ln \left( \frac{b}{x} \right)-\frac{1}{b}=0 \rightarrow x=e^{\left( \frac{1}{b} \right)}$$
But the answer should be ##x=\frac{b}{e}##

 

[SteamKing]

Have you tried rewriting ln (b / x) using the rules of logarithms before taking the derivative of the whole expression?

 

[SammyS]

Homework Statement

$$y=a\cdot x\cdot ln\left(\frac{b}{x}\right)$$
The derivative should be 0 (to maximize), what's x?

Homework Equations

$$(ln\:x)'=\frac{1}{x}$$
$$(x^a)'=ax^{(a-1)}$$
$$(uv)'=u'v+v'u$$

The Attempt at a Solution

$$\dot y=a \left[ ln \left( \frac{b}{x} \right)-x\frac{x}{b}x^{-2} \right]=a \left[ ln \left( \frac{b}{x} \right)-\frac{1}{b} \right]$$
$$\dot y=0 \rightarrow ln \left( \frac{b}{x} \right)-\frac{1}{b}=0 \rightarrow x=e^{\left( \frac{1}{b} \right)}$$
But the answer should be ##x=\frac{b}{e}##

You missed the Chain rule.

Also it may be make things easier to first expand the logarithm.

##\displaystyle \ \ln\left(\frac{b}{x}\right)=\ln(b)-\ln(x) \ ##

 

[Karol]

This question is from my post in physics, i re-wrote the constants in order not to confuse.
Yes, i used there the logarithm's rules:
$$y=a\cdot x\cdot ln\left(\frac{b}{x}\right)=a\cdot x\cdot (ln\:b-ln\:x)$$
$$\dot y=a\left[( ln\:b-ln\:x )-x\frac{1}{x} \right]=a(ln\:b-ln\:-1)=a\left[ ln\left( \frac{b}{x}\right)-1 \right]$$

 

[Karol]

You missed the Chain rule.

$$\left[x\cdot ln\left( \frac{b}{x} \right)\right]'=ln \left( \frac{b}{x} \right)+x\frac{x}{b}(-b)\frac{1}{x^2}=x\left[ ln\left( \frac{b}{x} \right)-1\right]$$
I think it's solved, i forgot the constant b, thanks

 

[SammyS]

$$\left[x\cdot ln\left( \frac{b}{x} \right)\right]'=ln \left( \frac{b}{x} \right)+x\frac{x}{b}(-1)\frac{1}{x^2}$$

What's the derivative of ##\displaystyle \ \frac{b}{x} \ ?\ ##

 

[Karol]

See my answer in #5, i solved, thanks

 

[SammyS]

See my answer in #5, i solved, thanks

That's incorrect.