Math – Exercise 4 (Geometry, Analysis)

Given

A pyramid with a square base.


Find

The minimum surface for a volume of 1000 cm3.


Solution


\(
\begin{array}{l}
V = \frac{1}{3}hx^2 = 1000cm^3 \\
S = x^2 + 2xh_\triangle, h_\triangle = \sqrt{0.5x^2+h^2}, S = x^2 + 2x\sqrt{0.5x^2+h^2} \\\\\\
h = \frac{3000cm^3}{x^2}, S = x^2 + 2x\sqrt{\frac{x^2}{2}+\frac{3000cm^3}{x^2}^2} \\
\frac{dS}{dx} = 2x + 2\sqrt{\frac{x^2}{2}+\frac{3000cm^3}{x^2}^2} + \frac{x(x-\frac{4*(3000cm^3)^2}{x^5})}{\sqrt{\frac{x^2}{2}+\frac{3000 cm^3}{x^2}^2}}, (\text{Product rule:}\ \frac{d}{dx}(u \cdot v) = \frac{du}{dx} \cdot v + u \cdot \frac{dv}{dx})\\
\frac{dS}{dx} = 0 \rightarrow x \approx 12.13cm,\ S_{min} \approx 683.78cm^2,\ h \approx 20.39cm \\\\\\
(V \dots \text{volume},\ S \dots \text{surface},\ h \dots \text{height},\ x \dots \text{side length})
\end{array}
\)


Answer

\(S_{min} \approx 683.78cm^2\)