More than one way to solve

By symmetry, the ellipse with the minimum area that passes through the four vertices of square $AB'C'D$ , with $B'(5, 0)$ and $C'(5, 5)$ , would be a circle.

The diameter of this circle is the same as the diagonal of the square, which is $d = 5\sqrt{2}$ , which means its radius is $r = \frac{5\sqrt{2}}{2}$ .

Stretching this square and circle horizontally by a factor of $2$ would give rectangle $ABCD$ and the ellipse with the minimum area, which has a semi-major axis of $a = 2r$ and a semi-minor axis of $b = r$ .

The area of this ellipse is $A = a b \pi = 2r^2 \pi = 2(\frac{5\sqrt{2}}{2})^2 \pi = 25\pi$ . Therefore, $n = \boxed{25}$ .

Denote the center of the ellipse as E. We know that the horizontal distance from E to any of A, B, C, or D is equal to 5. Likewise, we know that the vertical distance from E to any of A, B, C, or D is equal to 2.5. Therefore, point E is located at (0 + 5, 0 + 2.5) = (5, 2.5).

The equation for the ellipse cirumscribed around rectangle ABCD is: $\frac{\left(x-5\right)^{2}}{a^{2}}+\frac{\left(y-2.5\right)^{2}}{b^{2}}=1$ All points A, B, C, and D must satisfy the above equation. If we select any of these points, we will get the following relationship between $a$ and $b$ : $\frac{25}{a^{2}}+\frac{6.25}{b^{2}}=1$ And solving for $a$ gives us: $a = \sqrt{\frac{25}{1-\frac{6.25}{b^{2}}}} = \sqrt{\frac{25b^{2}}{b^{2}-6.25}} = 5 b \sqrt{\frac{1}{b^{2}-6.25}}$

The area of any ellipse is $A = \pi a b$ meaning the area of our ellipse is $A(b) = 5 \pi b^2 \sqrt{\frac{1}{b^{2}-6.25}}$ . We wish to find the value of $b$ which minimizes the ellipse's area - that is, we wish to find the values of $b$ where $A'(b) = 0$ and $(b, A(b))$ is a minimum. Using the product rule, $A'(b)$ is equal to: $A'(b) = \frac{10\pi b}{\left(b^{2}-6.25\right)^{\frac{1}{2}}}-\frac{5\pi b^{3}}{\left(b^{2}-6.25\right)^{\frac{3}{2}}}$ And further simplifying into a single term: $A'(b) = \frac{10\pi b\left(b^{2}-6.25\right)}{\left(b^{2}-6.25\right)^{\frac{3}{2}}}-\frac{5\pi b^{3}}{\left(b^{2}-6.25\right)^{\frac{3}{2}}}$ $A'(b).= \frac{5\pi b^{3}-6.25\cdot10\pi b}{\left(b^{2}-6.25\right)^{\frac{3}{2}}}$ Note that $A'(b)$ can only equal zero when the numerator $5\pi b^{3}-6.25\cdot10\pi b$ equals zero. $5\pi b^{3}-6.25\cdot10\pi b=0$ $b = \pm \sqrt{\frac{6.25\cdot10\pi}{5\pi}} = \pm \frac{5\sqrt{2}}{2}$ $b = 0$ would also be a valid solution to this equation, however, an ellipse with a vertical lengh of 0 would not be able to intersect with any of the corners of rectangle ABCD. Since there are only two critical points and $b = -\frac{5\sqrt{2}}{2}$ is unreasonable (for the same reason that $b = 0$ is unreasonable), we know that a value of $b = \frac{5\sqrt{2}}{2}$ yields the minimum value for the area of the ellipse. We know that $b$ yields the minimum area of the ellipse since it is impossible for the ellipse to have a maximum area (the ellipse can extend indefinitely in the y direction). Substituting this value for $b$ into $A(b)$ returns an area of $25 \pi$ meaning that the answer to the problem is $\boxed{25}$

The center of the ellipse is at the center of the rectangle $(5, 2.5)$ , hence its equation is

$\dfrac{(x - 5)^2}{a^2 } + \dfrac{(y - 2.5)^2 }{b^2} = 1$

Plugging in one of the four vertices, (from symmetry, one vertex is sufficient) we get,

$\dfrac{25}{a^2} + \dfrac{6.25}{b^2} = 1$

We're seeking the minimum of $ab$ , i.e. the maximum of $\dfrac{1}{ab}$ . Using AM-GM, we have

$1 = 2 (1/2) (\dfrac{25}{a^2} + \dfrac{6.25}{b^2}) \ge 2 \sqrt{ \dfrac{(25)(6.25)}{a^2 b^2} } = \sqrt{ \dfrac{ 25^2}{ a^2 b^2 } } = \dfrac{25}{ab}$

Hence, $a b \ge 25$ , and thus the minimum area of the circumscirbed ellipse is $25 \pi$ and $n = 25$ .