I am trying to find the maximum area of a triangle under some given constraints. I have generated a function to do this. If $x$ and $y$ are Real Numbers such that; $$x^2+y^2=1,$$ then the area of the triangle is given by the function; $$f(x,y)=x(1+y).$$
So, when is the value given by this function maximised (subject to $x^2+y^2=1$)?
We know that maximizing $f(x,y)$ is equivalent to maximizing $f(x,y)^2.$ We have; \begin{align*} f(x,y)^2 &=\left(x(1+y)\right)^2\\ &=x^2(1+y)^2\\ &=(1-y^2)(1+y)^2\\ &=(1-y^2)(y^2+2y+1)\\ &=-y^4-2y^3+2y+1. \end{align*}
Let this function be $g(y)=-y^4-2y^3+2y+1.$ We know that the derivative of this function is $-4y^3-6x^2+2,$ which is nonnegative for $y\leq \frac{1}{2},$ and negative for $y>\frac{1}{2}.$
Thus $g(y)$ incresases whenever $y\leq \frac{1}{2},$ and decreases when $y>\frac{1}{2},$ and acheives a maximum value $f(y)=\frac{27}{16}$ when $y=\frac{1}{2}.$
Thus the maximum value of $f(x,y)^2$ is $\frac{27}{16}$ when $y=\frac{1}{2}$ and $x=\sqrt{1-\dfrac{1}{4}}=\frac{\sqrt{3}}{2}.$
So, the maximum value of the original function $f(x,y)$ is; $$\sqrt{\frac{27}{16}}=\frac{3\sqrt{3}}{4}.$$
It is acheived when; $$\left(x,y\right)=\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right).$$