2
$\begingroup$

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?

  • 1
    Your formula for the height is not correct. If $a, b$ are the sides of the rectangle then $a^2+b^2 = 1$, by Pythagoras, since the radius is the diagonal of the rectangle.2012-03-28
  • 0
    @Thomas: The diagonal of the rectangle is the diameter, so $a^2 + b^2 = 2$.2012-03-28
  • 0
    $A(3/5)$ and $A(4/5)$ are equal because of symmetry: $(3/5)^2+(4/5)^2=1$2012-03-28
  • 0
    @martini: I think the diagonal of the rectangle is the *radius*2012-03-28
  • 0
    @Henry Thx, you and Thomas are right. I didn't read carefully.2012-03-28
  • 0
    @John Where is your problem with $\sqrt{\frac 12}$? As $\frac 9{25} < \frac 12 < \frac{16}{25}$ you have $\frac 35 < \sqrt{\frac 12} < \frac 45$.2012-03-28

1 Answers 1