Find maximum of $xy$ subject to $(x+1)^2 + y^2 = 4$ without using Calculus techniques Find maximum of $xy$ subject to $(x+1)^2 + y^2 = 4$.


It is easy if we apply Calculus techniques (e.g., derivatives, Lagrange multipliers, etc). However, the problem is assigned for students who have not yet learned Calculus. Is there a way to find the maximum without using Calculus techniques?

I have a few ideas:


Apply geometric means: $xy \leq \frac{x^2+y^2}{2} = -x + 3/2 $.
Change object function to $(xy)^2$, then we have $(xy)^2 = x^2(4-(x+1)^2)=x^2(-x^2 - 2x +3)$. That is, maximize  $x^2(-x^2 - 2x +3)$ on $(-3,1)$.
If you are deadset on a way without calculus, lets start with your idea $2$, i.e. how to maximize
$$p(x) = x^2y^2 = x^2\left(4-(x+1)^2 \right)= x^2(1-x)(3+x)$$

Now it is evident that this is a downward facing quartic with roots (with multiplicity) $-3, 0, 0, 1$, so the function is positive when $x \in (-3, 0)$ or $(0, 1)$.  It must have maxima in those intervals and one of them must be the global maximum we seek.

Case $1$: $x \in (-3, 0)$
$$p = (-x)(-x) (1-x)(3+x)$$
is a product of four positive terms, so by AM-GM, if the sum were a constant this would be maximized when all terms are equal.  Unfortunately the sum is not a constant, and the equality condition $(-x) = (1-x)$ is absurd.  Both these problems however can be addressed by appropriate scaling.  So consider maximizing for some real $\alpha, \beta &gt; 0$,
$$\alpha \beta p = (-x)(-x)(\alpha - \alpha x)(3\beta + \beta x)$$
Now the sum is a constant if we have $2+\alpha = \beta$ and for equality we must have $-x = \alpha -\alpha x = 3\beta + \beta x \implies \alpha = \frac14(\sqrt{33}-3), \beta = \frac14(5+\sqrt{33}), x = -\frac14(3+\sqrt{33})$

As all the conditions for AM-GM get satisfied here, the maximum in $(-3,0)$ is when $x = -\frac14(3+\sqrt{33}) = x_1$, say.

Case $2$: $x \in (0, 1)$
Following the same logic, we now have
$$\alpha \beta p = (x)(x)(\alpha-\alpha x)(3\beta+\beta x)$$
and the conditions give 
$2+\beta = \alpha, x = \alpha - \alpha x = 3\beta + \beta x \implies x_2 = \frac14(\sqrt{33}-3)$

All that is left is to see which among $x_1, x_2$ gives you the maximum, and it is easily seen $p(x_1) &gt; p(x_2)$, so the maximum is $\sqrt{p(x_1)} = \frac14\sqrt{\frac32(69+11\sqrt{33})} \approx 3.52$.
By looking at the plot, the polynomial $$y=(4-(x+1)^2)x^2$$ has three extrema.



In the vicinity of an extrema, the curve looks like a parabola. We can bring the vertex of the parabola on the axis by translating the independent variable, $x\leftarrow x-m$, giving

$$(4-(x-m+1)^2)(x-m)^2=(3m^2+2m^3-m^4)+(-6m-6m^2+4m^3)x+(3+6m-6m^2)x^2+(-2+4m)x^3-x^4.$$

This is achieved when the first degree term vanishes, i.e. when

$$-6m-6m^2+4m^3=0$$ or

$$m=0,m=\frac{3\pm\sqrt{33}}4.$$

The approximating parabolas are

$$y=(3m^2+2m^3-m^4)+(3+6m-6m^2)(x+m)^2,$$ giving the extrema values $$3m^2+2m^3-m^4.$$