# 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