SOLUTION: 1.A police officer is investigating a crime which occurred in a rectangular field next to a building. He wants to seal the three sides of the area around the scene with 300 m of y
Question 1008812: 1.A police officer is investigating a crime which occurred in a rectangular field next to a building. He wants to seal the three sides of the area around the scene with 300 m of yellow police tape. What is the maximum area that he can enclose and what are the dimensions of this area? (Hint: Draw a picture and introduce a variable x for the width…think about an expression for the length in terms of x) (6 marks)
PLEASE ANSWER FULL QUESTION GRADE 11 MATH!!! Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! He wants to seal the three sides of the area around the scene with 300 m of yellow police tape.
What is the maximum area that he can enclose and what are the dimensions of this area?
:
Let L = the length and x = the width of the rectangle
then because 3 sides are what is needed, therefore:
L + 2x = 300
L = (-2x+300), use this form for substitution
:
Area = L * x
Replace L
A = x(-2x+300)
A = -2x^2 + 300x
Maximum area occurs at the axis of symmetry, x = -b/(2a)
x =
x = +75 is width
then
L = -2(75) + 300
L = 150 is the length
:
Find the area: 150 * 75 = 11250 sq meters is max area
