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Let $a$ be the side of the equilateral triangle. The median of an equilateral triangle is given by $m = \frac{\sqrt{3}}{2}a$.
We are given that $\frac{dm}{dt} = 2\sqrt{3}$ cm/s. We need to find $\frac{da}{dt}$. Differentiating $m = \frac{\sqrt{3}}{2}a$ with respect to time $t$, we get: $$ \frac{dm}{dt} = \frac{\sqrt{3}}{2} \frac{da}{dt} $$
Substituting the given value of $\frac{dm}{dt}$, we have: $$ 2\sqrt{3} = \frac{\sqrt{3}}{2} \frac{da}{dt} $$ $$ \frac{da}{dt} = \frac{2 \cdot 2\sqrt{3}}{\sqrt{3}} = 4 $$ Thus, the rate at which the side is increasing is 4 cm/s.
Let the two numbers be $x$ and $y$. We are given that $x + y = 5$. We want to minimize $x^3 + y^3$.
Since $x + y = 5$, we can write $y = 5 - x$. Then, we want to minimize $f(x) = x^3 + (5-x)^3$.
$$ f(x) = x^3 + (5-x)^3 = x^3 + (125 - 75x + 15x^2 - x^3) = 15x^2 - 75x + 125 $$ $$ f'(x) = 30x - 75 $$
To find the critical points, set $f'(x) = 0$: $$ 30x - 75 = 0 $$ $$ x = \frac{75}{30} = \frac{5}{2} $$
$$ f''(x) = 30 > 0 $$ Since the second derivative is positive, $x = \frac{5}{2}$ corresponds to a minimum.
If $x = \frac{5}{2}$, then $y = 5 - x = 5 - \frac{5}{2} = \frac{5}{2}$.
The sum of the squares is $x^2 + y^2 = (\frac{5}{2})^2 + (\frac{5}{2})^2 = \frac{25}{4} + \frac{25}{4} = \frac{50}{4} = \frac{25}{2}$.
Final Answer: (a) 4 cm/s OR (b) 25/2
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