Assume that the potter orders $u$ units every $n$ days. It would be a waste to have units left over by the time she reorders, so the $u$ units must last exactly $n$ days. Since each day uses $5$ units, we must have $u = 5n$.

Let the material cost be $m$. The initial cost to the potter is thus $I = M + 5nm$. The warehousing cost can be calculated as follows. Let $t$ be the amount of days passed since the last order. Then the number of units left is given by $u(t) = 5n - 5t$. The warehousing cost per day is thus $w(t) = Ku(t) = 5K(n-t)$. The total warehousing cost is given by integration.$$W = \int_{0}^{n} w(t) \, \mathrm{d}t = \int_{0}^{n} 5K(n-t) \, \mathrm{d}t = \tfrac{5}{2}Kn^2.$$

Thus the cost over $n$ days is given by

$$I+W = M + 5nm + \tfrac{5}{2}Kn^2.$$

The cost per day, assuming she orders every $n$ days is

$$C(n, M, K) = 5m +M/n + \tfrac{5}{2}Kn.$$

To minimize the cost of clay, we seek to minimize $C$. We use the first derivative test:

$$\frac{\partial C}{\partial n} = 0 \implies \tfrac{5}{2}K - M/n^2 = 0,$$

which implies that

$$n = \sqrt{\frac{2M}{5K}}.$$

$\textbf{a)}$ If $M = 3000$ and $K = 10$, then she should order every $\sqrt{120} \approx 11$ days.

$\textbf{b)}$ We determine this by taking partial derivatives. First, for $M$, we have

$$\frac{\partial n}{\partial M} = \sqrt{\frac{1}{10MK}}.$$

So, increasing the delivery fee $M$ causes the number of days until the next order to increase, but this diminishes as $M$ and $K$ grow larger. This makes intuitive sense, as a larger delivery fee would make prevent one from excessively calling in orders.

For $K$, we have

$$\frac{\partial n}{\partial M} = -\sqrt{\frac{M}{10K^3}}.$$

So, increasing the warehousing fee $K$ causes the number of days until the next order to decrease. This effect diminishes as $K$ grows larger, but becomes more pronounced as $M$ grows larger. This makes intuitive sense, as a larger warehousing fee would prevent one from making large and infrequent orders, which would keep excessive amounts of clay in storage.