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Use the intended hot-ridge state and a linear law of thermal expansion about mantle density ${\rho }_m$ at $T_1$. Let $s$ be depth below the old seafloor. Its excess density isand its excess column mass is ${\int }_0^a\delta \rho ds={\rho }_m\alpha \Delta Ta/2$. Isostasy requires equal pressure at a compensation depth below both columns. Lowering the old seafloor by $h$ replaces rock by water, reducing column mass by $({\rho }_m-{\rho }_w)h$. Therefore the isostatic ridge elevation in a plate model isHere $\alpha \Delta T=0.052$; the result is a first-order estimate and neglects sediments, crustal-composition differences and flexural support. If one instead insists on the literal printed zero-age profile, its column is denser than the old plate and the formal relative elevation reverses sign, $-3.9\mathrm{km}$ to this order. That profile cannot describe the elevated hot ridge posited by the question.

Oceanic thermal subsidence. Newly formed oceanic lithosphere is hot and buoyant. Cooling increases its column density, causing subsidence by isostasy as it moves away from a ridge. The half-space cooling model has profile isWith volumetric thermal expansion coefficient $\alpha$, the excess column mass relative to hot mantle isCompensation with seawater density ${\rho }_w$ givesThus young-to-intermediate-age seafloor generally deepens approximately as the square root of age. At old ages, finite thermal-plate thickness and basal heat supply flatten this trend; indefinite half-space deepening is inappropriate. Sediments, variable crust, volcanic swells and dynamic topography also perturb bathymetry. Ocean-floor depth and heat flow jointly constrain cooling models.^[ref]

Effective elastic thickness. This is the thickness of a homogeneous elastic plate with the observed effective flexural rigidity, not necessarily the crustal or thermal lithospheric thickness:Infer it by modelling deflection around known loads, such as seamounts or trenches, or fitting gravity-topography admittance and gravity-topography coherence across wavelengths. A simple Fourier balance is $(D{k}^4+\Delta \rho g)w_k=q_k$ for a signed transverse load. The transition from buoyancy-dominated to bending-dominated behaviour constrains $D$; then assumed Young's modulus $E$ and Poisson's ratio $\nu$ give $T_e$.

Good inference includes load geometry and density, buried interfaces, sediment and internal loading, boundary conditions and data resolution. These can trade off with rigidity. Plastic yielding, faulting, mechanical layering and creep make $T_e$ an effective, timescale- and stress-dependent quantity. Older, colder lithosphere is often stronger, but age at loading and subsequent relaxation matter; a single universal age-to-$T_e$ conversion is unjustified.