Not sure how to differentiate implicitly using parametric equations...

Question

I am not sure (not taught before explicitly) how to apply implicit differentiation on parametric equations when I am solving the question posted below.

Question Two positive numbers $x$ and $y$ vary in such a way that $128x^2 - 16x^2y +1 = 0$. Find the minimum value of $x+y$.

My workings so far are:

I let $x = f(t), y = g(t)$ for some real $t$, then I perform implicit differentiation w.r.t $t$ on the equation (in the equation) to yield (can I perform implicit differentiation this way even for parametric equations, especially the term $x^2y$ above???)

$$256x\left(\frac{dx}{dt}\right) - 32xy\left(\frac{dx}{dt}\right) - 16x^2\left(\frac{dy}{dt}\right) = 0$$

Then, since $x>0$ as defined, and simplifying the equation above, I get that:

$$16\left(\frac{dx}{dt}\right) -2y\left(\frac{dx}{dt}\right) - x\left(\frac{dy}{dt}\right) = 0.$$

Then, (is this correct???) since if $x+y$ is minimum then $(\frac{dx}{dt}) + (\frac{dy}{dt}) = 0 \implies (\frac{dx}{dt}) = -(\frac{dy}{dt})$, then sub this into equation to get

$$16\left(\frac{dx}{dt}\right) -2y\left(\frac{dx}{dt}\right) + x \left(\frac{dx}{dt}\right) = 0.$$

From here, I am not sure how to proceed... Can anyone enlightened me on the "correctness" of my above workings as well as provide some hints on how to proceed? Thanks!

Btw, answer to this question is $8.75$.

Answer 1

It might be simpler to explicitly solve for the quantity you want to minimize, $x + y$. So let's call it $u$ and introduce it in your initial equation through $y = u - x$ :

$128x^2 - 16 x^2 (u -x) +1 = 0$

which yields $u = 8 + \frac{1}{16 x^2} + x$

Now let us differentiate $u$ w.r.t. $x$ (of course, if an implicit (ah ah) goal was to use implicit differentiation, that's not the way to go) :

$$\frac{du}{dx} = 1 - \frac{1}{8 x^3}$$

which shows that the minimum occurs for $x = \frac{1}{2}$, yielding $u = 8.75$

Answer 2

You've got $$ (16-2y+x)\frac{dx}{dt}=0. $$ This should hold regardless of the way in which $x$ varies with $t$; hence it is true when $dx/dt\ne0$, so we get $$ 16-2y+x=0\text{ or } y = 8 + \frac x 2. $$ So substitute $8+x/2$ for $y$ in the original equation $128x^2 - 16x^2y +1 = 0$: $$ 128x^2-16x^2\left( 8 + \frac x 2 \right) + 1 =0. $$ I'm not sure whether simply solving this numerically is the quickest way or not.