Extremum problem with multiple unknown variables

Question

(This is a homework question): We move with constant speed from A to B with $v_1$ and from B to C with $v_2$.

($v_1 < v_2$) and ($x <= b$) is known.

Please have a look at the image for clarity. Description of the image: We move (From a bird's eye view) diagonally from A to B and in a straight line from B to C.$$\begin{align*} A(0|a) \\ B(x|0) \\ C(b|0)\end{align*}$$

How do we choose x in dependence of $a, v_1,v_2$, so that we minimize the total amount of time needed to get to C?

What I have tried far:

$$t=\dfrac{s_{1}}{v_{1}}+\dfrac{s_{2}}{v_{2}}$$

$$s_{1}=\sqrt{a^{2}+x^{2}}$$

$$s_{2}=b-x$$

$$t=\dfrac{\sqrt{a^{2}+x^{2}}}{v_{1}}+\dfrac{b-x}{v_{2}}$$

And I am stuck. I think calculating the derivative and solving it for $0$ is not feasible right now and that maybe there is some way to get rid of more variables.

How do I need to proceed?

Thank You

Answer 1

$$t=\dfrac{\sqrt{a^{2}+x^{2}}}{v_{1}}+\dfrac{b-x}{v_{2}}$$ $$\implies \frac{dt}{dx} = \frac 1{v_1}\frac x{ \sqrt{a^{2}+x^{2}}} -\frac 1{v_2}$$ for min , must have $ \frac{dt}{dx} = 0$ $$\implies \frac 1{v_1}\frac x{ \sqrt{a^{2}+x^{2}}} =\frac 1{v_2}$$

now use my hint ( which is valid for $x \ge 0$) to solve for $x$ in terms of $a, b, v_1, v_2$.

It happens for this problem that $x$ does not depend on $b$ ( except for the condition $x \le b$ )