A rectangle with sides on the positive $x$-axis and positive $y$-axis is inscribed in the circle of radius $1$ with center at the origin. If $A(x)$ is the area of this rectangle for $\frac 35 \le x \le \frac 45$, so $A\colon \big[\frac 35, \frac 45\big] \to \mathbb R$, find those $x \in \big[\frac 35, \frac 45\big]$ where $A$ attains a maximum value and where $A$ attains a minimum value, or say no such values exist.
A graph of the problem looks like a circle with radius one centered on the origin, with a rectangle inside of it. The top-right corner of the rectangle stays in contact with the circle no matter what length is chosen, i.e. the length and width are related in that changing the length requires the width to be changed so that the top-right corner stays touching.
Here is what I have so far:
- Since $A$ is base $\cdot$ height, and the problem is wanting to find the base ($x$) that optimizes, the height needs to be expressed in terms of the base. Since the rectangle is in a circle, the height is going to be radius $-$ width, radius is 1, so that is $1 - $ width.
- So, $A = w(1 - w)$, or $A = w - w^2$, and A'(x) = 1 - 2w
- The root of the derivative A'(x) = 0 is, by my calculation, $\frac 12$
- $\frac 12$ is outside the interval
Am I correct in concluding that the maximum and minimum area for any $w$ on the interval are $A({\scriptstyle\frac 35})$ and $A({\scriptstyle\frac 45})$, respectively?
Edit:
I apologize, I'm having trouble with MathJaX here.
Corrected, $h = \sqrt{1 - w^2}$. So $A(b) = b \cdot \sqrt{1 - b^2}$. A'(b) = \frac{1 - 2b^2}{\sqrt{1 - b^2}}.
I still get the roots as $-\sqrt{\frac 12}$ and $+\sqrt{\frac12}$, can anyone see what I may be doing wrong?