A vertical column of height h carries a load Q at its

A vertical column of height $h$ carries a load $Q$ at its top. Find how the cross-sectional area must vary with position if, at every height, the cross-section is proportional to the total force (applied load plus weight of column above) acting there.

Step 1: Force as a function of position

Let $x$ measure distance down from the top, $A(x)$ the cross-sectional area, $\rho$ the material's density. The total force supported at $x$ is the applied load plus the weight of everything above it:

\[F(x)=Q+\int_0^x\rho g\,A(\xi)\,d\xi.\]

Step 2: Impose the proportionality condition

The problem's premise is $A(x)=kF(x)$ for some constant $k$ (area per unit force). Substituting:

\[F(x)=Q+k\rho g\int_0^xF(\xi)\,d\xi.\]

Step 3: Differentiate to remove the integral

\[\frac{dF}{dx}=k\rho g\,F(x).\]

Step 4: Solve the differential equation

Separating variables, $\dfrac{dF}{F}=k\rho g\,dx$, so $\ln F=k\rho g\,x+C$; the condition $F(0)=Q$ (just the applied load at the very top) gives $C=\ln Q$, hence

\[\boxed{F(x)=Qe^{k\rho gx}}\]

Step 5: Cross-sectional area

\[\boxed{A(x)=kF(x)=kQe^{k\rho gx}}\]

so the area (and, for a circular column, the radius $r(x)=\sqrt{A(x)/\pi}$) grows exponentially with depth. This exponential taper is exactly what the proportionality premise demands: the ratio of bottom-to-top cross-section is $A(h)/A(0)=e^{k\rho gh}$, a design profile distinct from a column of "uniform stress" (where area would instead be proportional to the constant allowable stress divided into the force).