SOLUTION: Officer Mark has 300 m of yellow tape to seal off the area of a crime scene. Using the tape, he must divide the area into four identical rectangular sections, side by side. What d

Algebra ->  Rational-functions -> SOLUTION: Officer Mark has 300 m of yellow tape to seal off the area of a crime scene. Using the tape, he must divide the area into four identical rectangular sections, side by side. What d      Log On


   

Question 1172937: Officer Mark has 300 m of yellow tape to seal off the area of a crime scene. Using the tape, he must divide the area
into four identical rectangular sections, side by side. What dimensions of the crime scene will maximize the area?
What is the maximum area that can be enclosed?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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Officer Mark has 300 m of yellow tape to seal off the area of a crime scene. Using the tape, he must divide the area into four identical rectangular sections, side by side.
What dimensions of the crime scene will maximize the area?
To divide into 4 areas, it requires 5 dividing sides and 2 lengths
2L + 5w = 300
simplify, divide by 2
L + 2.5w = 150
L = 150 - 2.5w
Area = L*w
replace L with (150-2.5w)
A = w(150-2.5w)
A = 150w - 2.5w^2
A quadratic equation where b=150 and a=-2.5
The max area occurs on the axis of symmetry, find that using x=-b/(2a)
w = %28-150%29%2F%282%2A-2.5%29
w = 30 meters is the width
Find the length
L = 150 - 2.5(30)
L = 75 meters is the length
:
What is the maximum area that can be enclosed?
75 * 30 = 2250 sq meters