Asked by T
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?
Answers
Answered by
bobpursley
Area=LW
perimeter=L+2W or L=300-2W
area=(300-2W)W=300W-2W^2
so at this point you can graph that Area vs W, (it is a parabola).
or you can look for the zeroes (300-2w)w, so zeros at w=0, w=150 which tells you the vertex is at w=75, so you can calculate the area at w=75. Or, you can go to calculus
Area=300W-2W^2
dArea/dw=300-4W=0 or W=75 at max area, then L=150, and area = ...LW
perimeter=L+2W or L=300-2W
area=(300-2W)W=300W-2W^2
so at this point you can graph that Area vs W, (it is a parabola).
or you can look for the zeroes (300-2w)w, so zeros at w=0, w=150 which tells you the vertex is at w=75, so you can calculate the area at w=75. Or, you can go to calculus
Area=300W-2W^2
dArea/dw=300-4W=0 or W=75 at max area, then L=150, and area = ...LW
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