# Use First Derivative to Minimize Area of Pyramid Optimization Problem

The first derivative is used to minimize the surface area of a pyramid with a square base.

## Problem

Below is shown a pyramid with square base, side length $$x$$, and height $$h$$. Find the value of $$x$$ so that the volume of the pyramid is 1000 cm 3 and its surface area is minimum.

Solution to Problem 1:

This problem has been solved graphically . Here we solve it more rigorously using the first derivative.
We first use the formula of
the volume of a pyramid to write the equation:
$\frac{1}{3} h x^2 = 1000$
The pyramid is made up of 4 triangles and a square (base)
The area of one triangle is given by
$$s = \dfrac{1}{2} H x$$
The slant height $$H$$ is given by
$$H = \sqrt{h^2 + \left(\frac{x}{2}\right)^2}$$
The surface area $$S$$ of the pyramid is given by the sum of the 4 areas of the 4 triangle and the area of the base $$x^2$$.
$$S = 4 (\dfrac{1}{2} H x) + x^2$$
$$= 2x \sqrt{h^2 + \left(\frac{x}{2}\right)^2} + x^2$$
Solve the equation $$\dfrac{1}{3} h x^2 = 1000$$ for $$h$$ to obtain:
$$h = \dfrac{3000}{x^2}$$
Substitute $$h$$ in the surface area formula by $$\dfrac{3000}{x^2}$$ to obtain a formula for $$S$$ in terms of $$x$$ ($$x$$ positive) only and rewrite it as follows:
$$S = 2x \sqrt{\left(\dfrac{3000}{x^2}\right)^2 + \left(\frac{x}{2}\right)^2} + x^2$$
Rewrite $$S$$ as
$$S = \sqrt{(36 \times 10^6 + x^6)} \cdot \dfrac{1}{x} + x^2$$
$$\quad = \dfrac{\sqrt{36 \times 10^6 + x^6}}{x} + x^2$$
Let constant $$k = 36 \times 10^6$$ and differentiate $$S$$ with respect to $$x$$.
$$\dfrac{dS}{dx} = \left[\dfrac{3x^6(k + x^6)^{-1/2} - (k + x^6)^{1/2}}{x^2}\right] + 2x$$
Multiply numerator and denominator by $$(k + x^6)^{1/2}$$ and simplify.
$$\dfrac{dS}{dx} = \dfrac{2x^6 - k}{x^2(k + x^6)^{1/2}} + 2x$$
A graph of $$\dfrac{dS}{dx}$$ is shown below. For $$x > 0$$, $$\dfrac{dS}{dx}$$ has a zero and is negative to the left of that zero and positive to the right of the zero. This means that $$S$$ has a minimum value that may be located by setting $$\dfrac{dS}{dx} = 0$$ and solve for $$x$$.

$\frac{2x^6 - k}{x^2(k + x^6)^{1/2}} + 2x = 0$
Rewrite as.
$\frac{2x^6 - k}{x^2} = -2x(k + x^6)^{1/2}$
Let $$u = x^3$$ and $$u^2 = x^6$$ and rewrite the above as follows:
$2u^2 - k = -2u(k + u^2)^{1/2}$
Square both sides:
$4u^4 + k^2 - 4ku^2 = 4u^2(k + u^2)$
Simplify to obtain:
$k^2 - 4ku^2 = 4ku^2$
Solve for $$u$$: ($$u$$ positive since $$x$$ positive)
$u = \sqrt{\frac{k}{8}}$
Finally solve for $$x$$ to obtain:
$x = \left(\sqrt{\frac{k}{8}}\right)^{1/3}$
Substitute $$k$$ by its value and calculate $$x$$:
$x = 12.8 \text{ cm}$ (rounded to 1 decimal place).
Below are shown the graphs of the surface area and its derivative.