The first derivative is used to minimize the surface area of a pyramid with a square base.
In what follows, \( C \) represents the constant of integration where applicable.
Problem
Below is shown a pyramid with a square base of side length \( x \) and height \( h \). Find the value of \( x \) so that the volume of the pyramid is \( 1000 \; \text{cm}^3 \) and its surface area is minimum.
Show Solution to Problem
This problem has been solved graphically. Here we solve it more rigorously using the first derivative.
We first use the formula for the volume of a pyramid to write the equation:
\[ \frac{1}{3} h x^2 = 1000 \]The pyramid consists of 4 triangular faces and 1 square base. The area of one triangular face is given by:
\[ s = \frac{1}{2} H x \]The slant height \( H \) is given by the Pythagorean theorem:
\[ H = \sqrt{h^2 + \left(\frac{x}{2}\right)^2} \]The total surface area \( S \) of the pyramid is the sum of the areas of the 4 triangles and the base area \( x^2 \):
\[ S = 4\left(\frac{1}{2} H x\right) + x^2 = 2x \sqrt{h^2 + \left(\frac{x}{2}\right)^2} + x^2 \]Solve the volume equation \( \dfrac{1}{3} h x^2 = 1000 \) for \( h \):
\[ h = \frac{3000}{x^2} \]Substitute \( h \) into the surface area formula to express \( S \) as a function of \( x \) alone:
\[ S(x) = 2x \sqrt{\left(\frac{3000}{x^2}\right)^2 + \frac{x^2}{4}} + x^2 = 2x \sqrt{\frac{9 \times 10^6}{x^4} + \frac{x^2}{4}} + x^2 \] \[ S(x) = 2x \sqrt{\frac{36 \times 10^6 + x^6}{4x^4}} + x^2 = 2x \cdot \frac{\sqrt{36 \times 10^6 + x^6}}{2x^2} + x^2 \] \[ S(x) = \frac{\sqrt{36 \times 10^6 + x^6}}{x} + x^2 \]Let constant \( k = 36 \times 10^6 \) and differentiate \( S \) with respect to \( x \):
\[ \frac{dS}{dx} = \frac{x \cdot \frac{1}{2}(k + x^6)^{-1/2}(6x^5) - \sqrt{k + x^6}}{x^2} + 2x \] \[ \frac{dS}{dx} = \frac{\frac{3x^6}{\sqrt{k + x^6}} - \sqrt{k + x^6}}{x^2} + 2x = \frac{3x^6 - (k + x^6)}{x^2 \sqrt{k + x^6}} + 2x = \frac{2x^6 - k}{x^2 \sqrt{k + x^6}} + 2x \]
For \( x > 0 \), \( \dfrac{dS}{dx} \) has a single zero, being negative to the left and positive to the right. This confirms that \( S \) has a minimum value located by setting \( \dfrac{dS}{dx} = 0 \):
\[ \frac{2x^6 - k}{x^2 \sqrt{k + x^6}} + 2x = 0 \implies \frac{2x^6 - k}{x^2} = -2x \sqrt{k + x^6} \]Let \( u = x^3 \) (so \( u^2 = x^6 \)) and substitute:
\[ \frac{2u^2 - k}{u^{2/3}} \quad \text{or simplifying directly:} \] \[ 2x^6 - k = -2x^3 \sqrt{k + x^6} \implies 2u^2 - k = -2u \sqrt{k + u^2} \]Square both sides:
\[ (2u^2 - k)^2 = 4u^2(k + u^2) \implies 4u^4 + k^2 - 4ku^2 = 4ku^2 + 4u^4 \] \[ k^2 - 4ku^2 = 4ku^2 \implies k^2 = 8ku^2 \implies u^2 = \frac{k}{8} \]Since \( u > 0 \):
\[ u = \sqrt{\frac{k}{8}} \implies x^3 = \sqrt{\frac{k}{8}} \implies x = \left(\frac{k}{8}\right)^{1/6} \]Substitute \( k = 36 \times 10^6 \):
\[ x = \left(\frac{36 \times 10^6}{8}\right)^{1/6} = (4,500,000)^{1/6} \approx 16.19 \; \text{cm} \](Note: Rounded to 1 decimal place as \( 16.2 \; \text{cm} \).)
