Answer
Assuming that $W$ remains constant and $I$ decreases by $1$, the estimated change in height is $\Delta H \simeq 0.02662$ meters.
Work Step by Step
We are given $I = \frac{W}{{{H^2}}}$. The partial derivatives are
$\frac{{\partial I}}{{\partial W}} = \frac{1}{{{H^2}}}$, ${\ \ \ }$ $\frac{{\partial I}}{{\partial H}} = - \frac{{2W}}{{{H^3}}}$
At $\left( {W,H} \right) = \left( {25,1.1} \right)$, we have
$\frac{{\partial I}}{{\partial W}}{|_{\left( {25,1.1} \right)}} = \frac{1}{{{{\left( {1.1} \right)}^2}}}$, ${\ \ \ }$ $\frac{{\partial I}}{{\partial H}}{|_{\left( {25,1.1} \right)}} = - \frac{{2\cdot25}}{{{{\left( {1.1} \right)}^3}}}$
Using the linear approximation, Eq. (5) the change in $I$ is estimated to be
$\Delta I \approx \frac{{\partial I}}{{\partial W}}{|_{\left( {25,1.1} \right)}}\Delta W + \frac{{\partial I}}{{\partial H}}{|_{\left( {25,1.1} \right)}}\Delta H$
$\Delta I \approx \frac{1}{{{{\left( {1.1} \right)}^2}}}\Delta W - \frac{{2\cdot25}}{{{{\left( {1.1} \right)}^3}}}\Delta H$
Assuming that $W$ remains constant and $I$ decreases by $1$, we have
$\Delta W = 0$, ${\ \ \ }$ $\Delta I = - 1$
Thus, the equation for $\Delta I$ becomes
$ - 1 \simeq - \frac{{2\cdot25}}{{{{\left( {1.1} \right)}^3}}}\Delta H$
$\Delta H \simeq 0.02662$
Thus, assuming that $W$ remains constant and $I$ decreases by $1$, the estimated change in height is $\Delta H \simeq 0.02662$ meters.
