Let the total cost function of a firm be given by:
$$TC(Q)= 16Q^3 - 72Q^2 + 446Q + 90$$
Find the level of production that minimises the marginal cost of production. (This is basically taking the first derivative of $TC$ then equating it to zero and finding the value of $Q$- this is the part I can't do).
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Taking the derivative of total cost with respect to $Q$ we obtain $$MC(Q)=\frac{\mathrm{d TC(Q)} }{\mathrm{d} Q}=48Q^{*2}-144Q^{*}+446=0$$
Then use the quadratic formula to solve for $Q^*$. This $Q^*$ will minimize the total cost but I guess you want the $Q$ that minimizes the marginal cost.
So taking the second derivative of the marginal cost we obtain $$96Q^{**}-144=0$$
Then solve for $Q^{**}$. This $Q^{**}$ will minimize your marginal cost.
No, if you set the first derivative equal to $0$ then you will find the point where the total cost is minimized. In order to minimize the marginal cost you should the second derivative equal to $0$.
The derivative of $TC$ with respect to $Q$ is equal to $$\frac{d}{dQ}TC(Q)=48Q^2-144Q+446=$$ and this is the marginal cost $MC$, i.e. $$MC(Q)=48Q^2-144Q+446$$ So, do not equate this to zero!(By the way, this has only imaginary roots). This is what you want to minimize! So take its derivative and proceed as you know!
The marginal cost is a convex function and therefore attains a minimum at the point which $\frac{d}{dQ}MC(Q)=0$. So $$\frac{d}{dQ}MC(Q)=96Q-144\overset{!}=0 \implies Q^*=\frac43$$ So, the marginal cost (not the total cost) is minimized at $Q=4/3=1.333$.