A cattle rancher wants to enclose a rectangular area and then divide it into six pens with fencing parallel to one side of the rectangle (see the figure below). There are 480 feet of fencing available to complete the job. What is the largest possible total area of the six pens?
Calculus - Damon, Monday, April 23, 2012 at 8:17pm
finally I get to use calculus for this problem instead of completing square!
length total = 6 x
width = y
fencing = 7 y + 12 x = 480
y = (480-12x)/7
A = 6 x y
A = 6 (x)(480-12 x)/7
A = (6/7) (480 x - 12 x^2 )
dA/dx = (6/7)(480 - 24 x) = for max
24 x = 480
x = 20
then y = 240/7 = 34.3 approx
A = 6 x y = 6 * 20 * 240/7 = 4114.3
Calculus - Damon, Monday, April 23, 2012 at 8:19pm
I have done many of these fencing problems tonight but most were in algebra where I had to find the vertex of the parabola by completing the square instead of taking the derivative and setting it to zero. Using Calculus makes it much much easier.