Optimization of $f(x,y)$ parametrizing the costraint

Question

I'm having problems understanding how parametrizing to optimize a two-variable function works. I have to optimize $$f(x,y) = x^2+y^2 + \frac{3}{2}x+1$$ with the costraint $4x^2+y^2-1=0$ . I tried to do firstly with Lagrange multiplier, and I found these points: $$\left(\pm\frac{1}{2},0\right), \left(\frac{1}{4},\pm\frac{\sqrt{3}}{2}\right)$$ But if I try parametrizing the function, for example, $f(x,\pm\sqrt{1-4x^2})$ I don't get all the points I need. Why this happens? I need to parametrize using $f\left( y,\pm\frac{\sqrt{1-y^2}}{2}\right)$ too. Why this happens?

Answer 1

We have the function: $$f(x,y) = \left(x+\frac{3}{4}\right)^2 + y^2 + \frac{7}{16}$$ This is a (imaginary?) circle with centre $C(-\frac{3}{4}, 0)$. The value of $f(x,y)$ at $P(x,y)$ depends on distance of $P$ from $C$ ($d^2 + 7/16$, where $d$ is distance of $P$ from $C$).

The other equation is an ellipse:

$$\frac{x^2}{1/4} + y^2 = 1$$

Thus you have to find such point $P$ on ellipse whose distance from $C$ is maximum.

For solving: Using your method, we have:

$$y^2 = 1-4x^2$$

Substitute in original equation and call this $g(x)$:

$$g(x) = 2-3x^2 + \frac{3}{2}x$$

$g(x)$ is maximum at $x = \frac{1}{4}$. This is easily verified from $g'(x)$.

Thus you get two points where $f(x,y)$ is maximum : $(\frac{1}{4}, \pm \frac{\sqrt{3}}{2})$

Answer 2

HINT: plug in $$y^2=1-4x^2$$ in your function, then it containes only the variable $x$