The maximum value of the two variables function given some constraints.

Question

Let:

$$f(x,y)=x^2+2y^2$$

It is required to obtain the maximum value of the above function subject to the constraint $$y-x^2+1=0.$$

I know how to maximize a function of two variables using the usual calculus method of finding the partial derivatives. But given a constraint, i have no idea how to proceed. One thing I tried is to obtain the value of $x^2$ from the constraint and substitute it to the two variable function. In that way, i get a one variable function which is entirely depending on $y$. Then, i tried to differentiate the obtained expression with respect to $y$. This is not giving me the desired result and i am stuck here. Any help would be very beneficial for me. Thanks.

Answer 1

Substituting $x^2$ of your constraint equation:

$f(y)= y+1+ 2y^2.$

This function is not bounded above .

Answer 2

for simplification you can consider the function $$h(y)=y+1+2y^2$$

Answer 3

$$x^2=y+1$$

and the objective function become $$f(y)=y+1+2y^2$$ where we require $y \geq -1$. Notice that $f(y)$ is a convex function.

$$f'(y)=4y+1$$

Hence the turning point is at $-\frac14$ and the function $y$ increases to $\infty$ after that.

In particular, let $(x,y)=(\sqrt{y+1},y)$ where $y>0$, then $f(x,y)=y+1+2y^2$ and it can be arbitarily large and the constraint is satisfied.

To understand this geometrically, plot the curve of $y=x^2-1$ and notice that the curve can go arbitrarily far away from the origin, i.e. $(x^2+y^2)+y^2$ can become arbitrarily large.