A small factory makes a metal alloy, and the manager studies the cost of making q tonnes in one day…

A small factory makes a metal alloy, and the manager studies the cost of making $q$ tonnes in one day. She knows the marginal cost, which is the extra cost of making one more tonne. It is $C'(q) = \dfrac{6q}{\sqrt{3q^2 + 16}}$ thousand rupees per tonne. She also knows the fixed cost, which is the cost when nothing is made. It is $20$ thousand rupees, so $C(0) = 20$. The manager notices that the top of the fraction is nearly the derivative of the expression under the root, so she plans a substitution with $u = 3q^2 + 16$. After she finds the total cost $C(q)$, she must fix the constant of integration with the fixed cost. She writes the answer as $C(q) = a\sqrt{3q^2 + 16} + K$. Then she wants two numbers for a day on which the factory makes $4$ tonnes: the total cost, and the marginal cost. Find $a$ and $K$, the total cost for $4$ tonnes in thousand rupees, and the marginal cost at $4$ tonnes in thousand rupees per tonne.