A rectangular prism with a square base can be described by its width $w$ and its height $h$. Let $s$ be the side length of the square base.

[asy] import three; unitsize(0.4 cm); currentprojection = perspective(4,2,2); draw(surface((1,1,1)--(3,1,1)--(3,3,1)--(1,3,1)--cycle),gray(0.7),nolight); draw(surface((1,1,-1)--(3,1,-1)--(3,3,-1)--(1,3,-1)--cycle),gray(0.7),nolight); draw(surface((3,2,1)--(4,2,0)--(3,2,-1)--(2,2,0)--cycle),gray(0.9),nolight); draw(surface((1,3,1)--(3,3,1)--(3,3,-1)--(1,3,-1)--cycle),white,nolight); draw(surface((3,1,1)--(3,3,1)--(3,3,-1)--(3,1,-1)--cycle),gray(0.8),nolight); draw((0,0,0)--(4,0,0)--(4,4,0)--(0,4,0)--cycle); draw((0,0,0)--(0,0,1)); draw((4,0,0)--(4,0,1)); draw((4,4,0)--(4,4,1)); draw((0,4,0)--(0,4,1)); draw((0,0,0)--(0,4,0)); draw((4,0,0)--(4,4,0)); draw((0,0,1)--(0,4,1)); draw((4,0,1)--(4,4,1)); draw((0,4,1)--(4,4,1)--(4,0,1)--(0,0,1)); draw((4,2,0)--(4,2,1)); draw((3,2,-1)--(3,2,1)); draw((2,2,0)--(2,2,1)); label("$w$", (2,0,0), SW); label("$s$", (0,2,0), W); label("$h$", (4,2,0), E); [/asy]

Let's imagine looking at the prism "from above", viewing the square base. The diagonal of the square base is also the diameter of the inscribed sphere.

[asy] size(5cm); defaultpen(linewidth(0.4)); pair A, B, C, D, E, F, G, H, I, J, K, L; E = (1,1); F = (3,3); A = (1,3); I = (0,2.1); H = (4,2.1); D = (1,2+0.3); J = (3,2+0.3); G = (2,2.7); K = (1.4,2.1); L = (2.6,2.1); B = (0.25,2.9); C = (1.75,1.95); draw((0,0)--(0,4)--(4,4)--(4,0)--cycle); draw(B--I--H--J--G--cycle); draw(I--D--J,dashed); draw(G--(2,0),dashed); label("$w$", (2,0), S); label("$s$", (0,2), W); label("$h$", (4,2), E); label("$s/\sqrt{2}$", midpoint(I--G), W); label("$s/\sqrt{2}$", midpoint(G--H), E,fontsize(8)); label("$5$", (2,0), S); label("$5$", (2,2), NE); label("$s/\sqrt{2}$", (0,1.05),W); label("$s/\sqrt{2}$", (4,1.05),E); [/asy]

Hence $s^2 + h^2 = 5^2$. From the Pythagorean Theorem on the right triangle, we can see that $(s/\sqrt{2})^2 + (w/2)^2 = (s/2)^2$, so $\frac{w^2}{4} + \frac{s^2}{2} = \frac{s^2}{4}$.

Thus, $w^2 = \frac{s^2}{2}$, and $w=\frac{s}{\sqrt2}$. We substitute this into the volume formula $V = s^2h$ to get \[V = \frac{s^2}{2} \cdot h = \frac12 sh^2\]since $h = \sqrt{5^2-s^2}$. Notice that $V$ is a product of three variables but only two dimensions are involved. Since $h=\sqrt{5^2-s^2} > 0$ for $0 < s < 5$, $V$ is increasing in $h$ as long as $s$ remains constant. Since $0 < s < 5$, we take out a factor of $\frac 12s$ and look for the maximum value of \[A = sh^2 = s(\sqrt{5^2-s^2})^2 = 25s - 2s^3.\]The maximum value of $A$ will occur when $A$ is at a maximum and we have $1 = \frac{dA}{ds} = 25-6s^2$. Thus, $s^2 = \frac{25}{6}$, and $s=\frac{5}{\sqrt{6}}$. By substituting back, we get $h = \sqrt{5^2-s^2} = \frac{5}{\sqrt{6}}$, and \[w = \frac{s}{\sqrt2} = \frac{5}{\sqrt{12}} = \frac{5\sqrt2}2.\]Therefore, the dimensions of the prism are $\boxed{\frac{5}{\sqrt6} \times \frac{5\sqrt2}{2} \times \frac{5}{\sqrt6}}$.

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