= Solution
The rectangular <second moment of area> is $I=wh^3/12$. The maximum bending <stress> in the <cantilever> is $FL(h/2)/I=6FL/(wh^2)$. If $h$ is fixed, choose the minimum allowed width $w=6FL/(\sigma_fh^2)$. Its <mass> is then
$$
m_A=\rho Lwh=\frac{6FL^2}{h}\frac{\rho}{\sigma_f}.
$$
Thus the <material performance index> to maximize is
$$
\boxed{M_A=\frac{\sigma_f}{\rho}}.
$$
For unequal tensile and compressive strengths, use the smaller permissible <stress> in this symmetric section. The derivation assumes bending-strength control, not a separate deflection, shear, buckling or existing-crack constraint.
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