Optimization of a Marker
Sam Montgomery
Original Marker
r = radius of marker = 0.25 in
h = height of marker = 5.5 in
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.