= Maximum lift-to-drag ratio
{title2=$(L/D)_{\max}$}
The \b[maximum lift-to-drag ratio] is the largest attainable ratio of positive <lift> to <drag> within stated operating conditions. It depends on both the <drag polar> and the range of <lift coefficients> the vehicle can actually attain.
For the positive <parabolic drag polar> $C_D=C_{D0}+kC_L^2$, a short proof uses the <arithmetic-geometric mean inequality>:
$$
C_{D0}+kC_L^2\ge2\sqrt{C_{D0}k}\,C_L,\qquad C_L>0.
$$
Consequently
$$
\boxed{(L/D)_{\max}=\frac1{2\sqrt{C_{D0}k}},\qquad C_L^*=\sqrt{C_{D0}/k},}
$$
provided that coefficient is attainable. Equality requires equal modeled zero-lift and lift-dependent drag contributions. This bound supplies a global argument, rather than merely locating a stationary point by <differentiation>.
A <lift-constrained maximum lift-to-drag ratio> can be lower. With $C_{D0}=0.020$, $k=0.050$ and allowed $0<C_L\le0.50$, the unconstrained optimum $C_L^*\simeq0.6325$ is excluded. Since the ratio increases up to that point, the constrained optimum is at $C_L=0.50$, giving \b[$L/D=15.3846$], below the unconstrained $15.8114$. A stall-limited interval should therefore be applied before claiming the fitted optimum is reachable.
<Parameter bounds on maximum lift-to-drag ratio> follow from monotonicity. If $C_{D0}\in[c_-,c_+]$ and $k\in[k_-,k_+]$ are positive intervals, then the unconstrained maximum lies in
$$
\left[\frac1{2\sqrt{c_+k_+}},\frac1{2\sqrt{c_-k_-}}\right].
$$
This <interval arithmetic> bound is conservative when the fitted parameters are correlated and not all interval combinations are jointly possible. It quantifies uncertainty in the assumed polar; it does not include failure of the polar model itself.
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