Text of the problem: A rectangle is inscribed in a semicircle and the radius is 1. The bas of the rectangle is x. Write an expression for the rectangle perimeter and determine the value of x that gives the highest possible perimeter. Also,, what is the highest perimeter?
Trigonometry is not to be used.
My Attempt:
the coordinates of the upper right corner will be $(\frac{x}{2}, \sqrt{1-\frac{x^2}{4}})$ and the area is then $$ A= x\sqrt{1-\frac{x^2}{4}} $$ and $$ \frac{dA}{dx} = \sqrt{1-\frac{x^2}{4}} - \frac{x^2}{4\sqrt{1-\frac{x^2}{4}}} $$ Solvine $\frac{dA}{dx} = 0$ will give the optimal $x$. The trick for that is to note that $x=1$ gives a rectangle that is just a line (the diameter of the semicircle) so we can assume that is not the case, and multiply through by $\sqrt{1-\frac{x^2}{4}}$. $$ 1-\frac{x^2}{4} = \frac{x^2}{4}\\ \frac{x^2}{4}=\frac12\\ x= \sqrt{2}\\ A = \sqrt{2}\sqrt{\frac12} = 1 $$
Let’s restate the problem slightly. Your figure is clearly symmetrical with respect to the perpendicular bisector of the diameter, so we can just look at the right-hand half of the picture. Coordinatize, so that the picture now lies in first quadrant of the plane, where the circle has equation $x^2+y^2=1$, and the perimeter of the (half-) rectangle is now $x+y$.
On the circle $x^2+y^2=1$, you’re maximizing $x+y$, a function whose level curves are the lines of slope $-1$. Certainly the line in this family that’s highest but still hits the circle is the one where $x+y=\sqrt2$, which hits at the $45^\circ$-point on the circle, $(\frac{\sqrt2}2,\frac{\sqrt2}2)$.
There’s your solution: In terms of Cartesian coordinates, the formula for the perimeter is $2(x+y)$, but with $y$ replaced by $\sqrt{1-x^2}$, the best value of the $x$-coordinate is $\frac{\sqrt2}2$, but for your $x$, it’s $\sqrt2$, and the max perimeter is $2\sqrt2$.