Proof. For fixed \(\boldsymbol \phi \) the model is linear with model matrix \(\mathbf {G}(\boldsymbol \phi )\), so the minimum over \(\boldsymbol \beta \) is the least-squares fit and the minimum value is the residual sum of squares \(\mathbf {y}^\top (\mathbf {I} - \mathbf {H})\mathbf {y}\) (Theorem 3.4.2). Then \(\min _{\boldsymbol \beta ,\boldsymbol \phi }S = \min _{\boldsymbol \phi }\min _{\boldsymbol \beta }S = \min _{\boldsymbol \phi }S_p(\boldsymbol \phi )\). With one column \(\mathbf {g}\), \(\mathbf {y}^\top \mathbf {Hy} = (\mathbf {g}^\top \mathbf {y})^2/\mathbf {g}^\top \mathbf {g}\). □

Solution. Lineweaver–Burk. Regressing \(1/y\) on \(1/x\) (with \(1/x = 2, 1, 0.5, 0.25, 0.125, 0.0625\) and \(1/y = 0.4348, 0.3226, 0.1923, 0.1563, 0.1220, 0.1136\)) gives \(1/y = 0.1115 + 0.1708(1/x)\), so \[ \tilde \theta _1 = \frac {1}{0.1115} = 8.968, \qquad \tilde \theta _2 = \frac {0.1708}{0.1115} = 1.532 . \] On the original scale these leave \(S = 1.057\).

Least squares. By Theorem 12.5.1, with \(g_i = x_i/(\theta _2 + x_i)\), \(\hat \theta _1(\theta _2) = \sum g_iy_i/\sum g_i^2\). A few trial values give

so the minimum is near \(\theta _2 = 2\). Gauss–Newton started from the Lineweaver–Burk values converges in four steps to \[ \hat \theta _1 = 9.938, \quad \hat \theta _2 = 1.958, \qquad S(\hat {\boldsymbol \theta }) = 0.300 , \] with standard errors \(0.328\) and \(0.208\) from \(s^2 = 0.300/4 = 0.075\). The fitted rates are \(2.02\), \(3.36\), \(5.02\), \(6.67\), \(7.98\), \(8.86\).

Comparison. The Lineweaver–Burk estimates leave a residual sum of squares three and a half times as large, and they lie two to three standard errors from the least-squares estimates. The reason is the transformation: the smallest rates, at the smallest concentrations, become the largest values of \(1/y\) and \(1/x\), and their errors are magnified most, so they dominate the linear fit. The transformation is useful for a starting value and for a plot, but not for the final estimates.