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While studying, I came upon this problem: "Find the maximum area of a rectangle circumscribed about a fixed rectangle of length 8 and width 4." I looked at the answer key, which showed that the maximum area possible was 72 inches squared. Also, it stated to use cosine and sine functions to solve the problem. However, I do not get how to apply these two functions to this problem. Could someone explain how to achieve this answer?

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Two vertices of the circumscribed rectangle are located on the semicircles centered at $O_1$ and $O_2$:

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The area of a rectangle $S$, circumscribed around the rectangle $ABCD$, consists of the fixed part $S_{ABCD}=8\cdot 4=32$ plus doubled sum of $S_{CED}$ and $S_{ADF}$, controlled by the angle $\phi$. Also, $\triangle CED$ is congruent to $\triangle ADF$. With the base of $\triangle CED$ fixed, the maximum of the area is reached when the height $|EG|$ is maximal, that is, $|EG|=r_1=\frac12|CD|=4$, when $\phi=\frac{\pi}{4}$.

$S_{\max}=8\cdot4+2(\frac12 8\cdot4+\frac12 4\cdot2)=72.$

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Hint: Drawing a picture is usually helpful. So draw a rectangle centered at the origin, and draw a ray emanating from the origin at an angle $\theta$ from the positive $x$-axis (for simplicity, take $0<\theta<\pi/2$, and take $\theta$ just slightly less than $\pi/2$ so the ray is just a little less than vertical).

Now draw the circumscribed rectangle whose sides are parallel/perpendicular to this ray (two will be parallel, two perpendicular). The interior of the circumscribed rectangle contains the original rectangle and four right triangles. The angles in these right triangles will be $\pi/2$, $\theta$, and $\pi/2-\theta$. The hypotenuse of each of these right triangles is a side of the original triangle. You can determine area of each of the right triangles as $\frac12\times$base$\times$height, and the base and height of each of them can be determined using the hypotenuse (a side of the original rectangle) and the sine or cosine (as appropriate according to your drawing) of one of the angles ($\theta$ or $\pi/2 - \theta$ as appropriate according to your drawing).

Now find which $\theta$ produces the greatest total area.

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In the figure below, the black rectangle is circumscribed around the blue rectangle:

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The edges of the two rectangles make an angle $\theta$, as shown in the figure.

Now see if you can tell what the lengths of $p$, $q$, $r$, and $s$ are in this figure, and if you can see what that has to do with the cosine and sine functions.

(Note that if you draw any blue rectangle and circumscribe a rectangle around it, the sizes of the two rectangles do not change when you rotate the entire figure in the plane. I have chosen to rotate the figure so that the black rectangle is "straight", both because it was easier to draw this way and because I think it helps in understanding the hint above. But if you don't like the orientation of this figure, rotate it again so that the blue rectangle is back exactly where it was in the first place; the lengths of the labeled segments will still be the same as they are now.)

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The circumscribed rectangle is the union of the original $a\times b$-rectangle and four similar right triangles with hypotenuses $a$, resp. $b$. The total area is maximal when one of these triangles, and therewith all of them, have maximal area, and this is the case when they are all isosceles.

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