Use First Derivative to Minimize Area of Pyramid
The first derivative is used to minimize the surface area of a pyramid with a square base.
Problem 1: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:
(1 / 3) h x^{ 2} = 1000

We now use the the formula of the surface area of a pyramid to write a formula for the surface area S of the given pyramid. In this problem we have a square pyramid, hence:
S = x * sqrt [ h^{ 2} + (x/2)^{ 2} ] + x * sqrt [ h^{ 2} + (x/2)^{ 2} ] + x * x
= 2 x sqrt [ h^{ 2} + (x/2)^{ 2} ] + x^{ 2}

Solve the equation (1 / 3) h x^{ 2} = 1000 for h to obtain:
h = 3000 / x^{ 2}

Substitute h in the surface area formula by 3000 / x^{ 2} to obtain a formula for S in terms of x (x positive) only and rewrite it as follows:
S = sqrt [ (36 10^{ 6} + x^{ 6}) ] / x + x^{ 2}

Let constant k = 36 10^{ 6} and differentiate S with respect to x.
dS / dx = [ 3 x ^{ 6} (k + x^{ 6})^{ 1/2}  (k + x^{ 6})^{ 1/2} ] / x^{ 2} + 2 x

Multiply numerator and denominator by (k + x^{ 6})^{ 1/2} and simplify.
dS / dx = [ 2 x ^{ 6}  k ] / [ x^{ 2} (k + x^{ 6})^{ 1/2} ] + 2x

A graph of dS / dx is shown below. For x > 0, 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 dS / dx = 0 and solve for x.
[ 2 x ^{ 6}  k ] / [ x^{ 2} (k + x^{ 6})^{ 1/2} ] + 2 x = 0

Rewrite as.
[ 2 x ^{ 6}  k ] =  2 x^{ 3} (k + x^{ 6})^{ 1/2}

Let u = x^{ 3} and u^{ 2} = x^{ 6} and rewrite the above as follows:
[ 2 u ^{ 2}  k ] =  2 u (k + u^{ 2})^{ 1/2}

Square both sides:
4 u ^{ 4} + k^{ 2}  4k u ^{ 2} = 4 u^{ 2} (k + u^{ 2})

Simplify to obtain:
k^{ 2}  4k u ^{ 2} = 4 k u^{ 2} k

Solve for u: (u positive since x positive)
u = sqrt [ k / 8 ]

Finally solve for x to obtain:
x = [ sqrt [ k / 8 ] ] ^{ 1/3}

Sustitute k by its value and calculate x:
x = 12.8 cm (rounded to 1 decimal place).

Below are shown the graphs of the surface area and its derivative.
More references on
calculus problems

