Remove grain before growth and retain the exact supply bound

Without laboratory supply, the year-end stock is $10000(1.02)^{12}\approx12682.41795$, or $12700$ bushels to 3 s.f. as printed.

Put $q=1.02$ and $V_0=10000$. With supply $100$:

Thus $V_3=10000q^3-100(q+q^2+q^3)=10299.9192\approx10299.9$ bushels.

For supply x, $V_m=q(V_{m-1}-x)$. January, February and March give $qV_0-qx$, $q^2V_0-x(q^2+q)$ and $q^3V_0-x(q^3+q^2+q)$. In December, $V_{12}=10000q^{12}-xq(q^{12}-1)/(q-1)$. The source variable-x table has a stray $100$ in one February start entry; the recurrence and general formula use x throughout.

Nonnegative December stock requires

$x\le\frac{10000q^{12}(q-1)}{q(q^{12}-1)}\approx927.0548689.$

The corresponding bound for month m decreases as m increases, so December is the tightest bound and all earlier withdrawals are feasible.

The source reports 927 bushels to 3 s.f. This is also the maximum if supply is restricted to whole bushels. If fractional bushels are allowed, the exact expression above is the actual continuous maximum; it must not be replaced by an unqualified exact integer bound.