15. Let the base of the pyramid $M-ABCD$ be a square, and $MA=MD, MA \perp AB$. If the area of $\triangle AMD$ is 1, find the radius of the largest sphere that can fit inside this pyramid.
When the sphere $O$ is tangent to the sides $M A D, M B C$, and the base $A B C D$, let $E, F$ be the midpoints of $A D, B C$ respectively. Since $A B \perp$ plane $M A D \Rightarrow \triangle M E F$ is a right triangle. At this time, the radius of the sphere $O$ is equal to the inradius $r$ of $\triangle M E F$.
As shown in the figure, let $A D=E F=a \Rightarrow M E=\frac{2}{a}$
$\Rightarrow r=\frac{M E+E F-M F}{2}$
$=\frac{a+\frac{2}{a}-\sqrt{a^{2}+\frac{4}{a^{2}}}}{2}=\frac{2}{a+\frac{2}{a}+\sqrt{a^{2}+\frac{4}{a^{2}}}}$
$\leqslant \frac{2}{2 \sqrt{2}+2}=\sqrt{2}-1$, equality holds when $a=\frac{2}{a} \Rightarrow a=\sqrt{2}$.
At this point, draw $O P \perp$ plane $M A B, O G \perp M E$ at $G, G Q \perp M A$ at $Q$, and connect $P Q$, then $G Q \perp$ plane $M A B \Rightarrow G Q \stackrel{\Perp}{=} O P$.
By $\triangle M Q G \sim \triangle M E A \Rightarrow \frac{Q G}{E A}=\frac{M G}{M A}$, and $E A=\frac{\sqrt{2}}{2}, M G=1, M A=\frac{\sqrt{10}}{2}$,
Thus, $O P=Q G=\frac{\sqrt{5}}{5}>\sqrt{2}-1$, which means the sphere $O$ is inside the pyramid $M-A B C D$. In summary, the maximum radius of the sphere that can fit inside this pyramid is $\sqrt{2}-1$.