Optimization of a Marker

Sam Montgomery

Original Marker


r = radius of marker = 0.25 in

h = height of marker = 5.5 in



surface area of a cylinder = 2πrh+2πr².

2π(.25)(5.5) + 2π(.25)² = 9.03 in²


volume of a cylinder = πr²h

π(.635)²(13.97) = 1.08 in³

Solve for the Variable h

2πrh + 2πr² = 9.03

2πrh = 9.03 - 2πr²

h = (9.03 - 2πr²)/2πr

Plug H into Volume Formula

volume of a cylinder = πr²h

h = (9.03 - 2πr²)/2πr


V = πr²((9.03 - 2πr²)/2πr)

= (9.03πr² - 2π²r⁴)/2πr

= (9.03r - 2πr³)/2

Take the Derivative of the Volume Formula

V = (9.03r - 2πr³)/2

V' = ((2)(9.03 - 6πr²) - (0)(9.03r - 2πr³))/2²

= (18.06 - 12πr²)/4

Set the Derivative Equal to Zero

0 = (18.06 - 12πr²)/4

0 = 18.06 - 12πr²

12πr² = 18.06

r² = 0.479

r = 0.6291

Plug R into Height Formula

h = (9.03 - 2πr²)/2πr

= (9.03 - 2π(.6921)²/2π(.6921)

= (9.03 - 3.01)/4.35

= 6.02/4.35

= 1.383

Optimal Marker

Height = 1.383 inches

Radius = 0.6921 inches

Volume = 2.08 inches³


Since the original volume of 1.08 divided by the optimal volume of 2.08 is about .519, the original marker's volume is about 51.9% optimized.