Finding the minimum of a function

Question

Let a and b be positve real numbers: \begin{align} f(a,b) = \dfrac{4a+b}{2a} + \dfrac{4a-3b}{b} \end{align}

Where is this function at a minimum?
What is this functions minimum value?

Answer: at a minimum at x = $2\sqrt2$, Minimum value: y = $2\sqrt2 -1$

I managed to take the derivative with respect to a, by first multiplying the denominators to get a common denominator. That gave me $$(4ab+b^2+8a^2-6ba)/2ab$$ After simplifying, I got $$(-2ab+b^2+8a^2)/2ab$$ I then took the derivative and got it all the way down to $$b' = (-16a+2b)/(2a+2b)$$ After simplifying a little bit more and setting the numerator to 0, I tried to get the critical points when I got $$b =8a$$. I realize that if I take the integral here, I will get the answer $$b = 2\sqrt2$$, but what I want to ask is, why would I take the integral there?

Maybe I'm just misunderstanding the question?

All help is appreciated! Thank you!

Answer 1

Hint

\begin{align} \frac{4a + b}{2a} + \frac{4a - 3b}{b} &= 2 + \frac{b}{2a} + \frac{4a}{b} - 3, \; a, b \ne 0 \\ & = \frac{b}{2a} + \frac{4a}{b} - 1\\ & = \frac{1}{2}\left(\frac{1}{\left(\frac{a}{b}\right)}\right) + 4\left(\frac{a}{b}\right) - 1\\ & = \frac{1}{2}\left(\frac{1}{x}\right) + 4x - 1 \end{align}

Answer 2

The expression can be simplified down to $$(2+\frac{b}{2a})+(\frac{4a}{b}-3)= \frac{b}{2a}+ \frac{4a}{b}-1$$
By the AM-GM inequality, $$\frac{\frac{b}{2a}+ \frac{4a}{b}}{2}\geq \sqrt{\frac{b}{2a}\cdot \frac{4a}{b}}=\sqrt2$$
So the minimum value of the entire expression is $2 \sqrt2-1$. This happens when $\frac{b}{2a}= \frac{4a}{b}$ or $b=2\sqrt2 a, $because a,b are positive.