By how much does the volume of a regular hexagonal

By how much does the volume of a regular hexagonal pyramid with lateral edge $l = 5\ \text{cm}$ increase when the base edge grows from $a = 4\ \text{cm}$ to $a = 4.1\ \text{cm}$?

Step 1: The volume as a function of the base edge

The base is a regular hexagon of side $a$, with area $\dfrac{3\sqrt3}{2}a^{2}$. In a regular hexagon the distance from the centre to a vertex equals the side, so the height satisfies $h^{2}+a^{2} = l^{2}$:

\[ h = \sqrt{25-a^{2}}, \qquad V(a) = \frac13\cdot\frac{3\sqrt3}{2}a^{2}\sqrt{25-a^{2}} = \frac{\sqrt3}{2}\,a^{2}\sqrt{25-a^{2}}. \tag{1} \]

The formula is valid for $0\lt a\lt 5$; as $a\to5$ the pyramid flattens to zero height.

Step 2: Differentiate

\[ V'(a) = \frac{\sqrt3}{2}\left[2a\sqrt{25-a^{2}}-\frac{a^{3}}{\sqrt{25-a^{2}}}\right] = \frac{\sqrt3}{2}\cdot\frac{2a\bigl(25-a^{2}\bigr)-a^{3}}{\sqrt{25-a^{2}}} = \frac{\sqrt3}{2}\cdot\frac{a\bigl(50-3a^{2}\bigr)}{\sqrt{25-a^{2}}}. \tag{2} \]

Step 3: Evaluate at $a = 4$

\[ \sqrt{25-16} = 3, \qquad 50-3\cdot16 = 2, \]

\[ V'(4) = \frac{\sqrt3}{2}\cdot\frac{4\cdot 2}{3} = \frac{4\sqrt3}{3} \approx 2.30940. \]

Step 4: The differential

\[ \Delta V \approx V'(4)\cdot 0.1 = \frac{4\sqrt3}{30} = \frac{2\sqrt3}{15} \approx 0.2309\ \text{cm}^{3}. \]

Step 5: The true increment — and why it differs

The exact values from (1) are

\[ V(4) = \frac{\sqrt3}{2}\cdot16\cdot3 = 24\sqrt3 \approx 41.5692, \qquad V(4.1) = \frac{\sqrt3}{2}\cdot16.81\cdot\sqrt{8.19} \approx 41.6620, \]

\[ \Delta V_{\text{exact}} \approx 0.0928\ \text{cm}^{3}, \]

only two-fifths of the estimate. The explanation is in formula (2): the factor $50-3a^{2}$ vanishes at

\[ a_{\max} = \sqrt{\frac{50}{3}} = \frac{5\sqrt6}{3} \approx 4.0825, \]

which lies inside the interval $[4,\,4.1]$. The volume rises to a maximum there and has already started to fall by $a = 4.1$:

\[ V\!\left(\frac{5\sqrt6}{3}\right) = \frac{125}{3} \approx 41.6667, \qquad V'(4.1) \approx -0.5335. \]

A tangent line drawn at $a = 4$ cannot possibly follow a curve that turns over within the step.

Step 6: Verification

\[V'(4) = +2.3094 \;\;\text{but}\;\; V'(4.1) = -0.5335 \quad\text{— the derivative changes sign in the interval}\]

\[V(4) = 41.5692 \;\lt \; V(4.1) = 41.6620 \;\lt \; V_{\max} = \frac{125}{3} = 41.6667 \;\;\checkmark\]

\[\text{Smaller steps converge: } \Delta a=0.01 \Rightarrow 0.021781 \text{ vs } 0.023094\ (6.0\%);\ \Delta a=0.001 \Rightarrow 0.0022963 \text{ vs } 0.0023094\ (0.57\%) \;\;\checkmark\]