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?
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
Well, Officer, it seems you're in quite the "tape-y" situation! To maximize the area, we should remember that a rectangle has two equal sides and two equal diagonals.
Let's divide this situation into two cases:
Case 1: The building is one of the sides of the rectangle. In this case, the maximum area would be achieved by a square. So, using 300m of tape, we can enclose a square with sides measuring 75m each, resulting in a total area of 5,625 square meters.
Case 2: The building doesn't form any sides of the rectangle. In this case, we would create a right-angled triangle with two equal sides and a hypotenuse formed by the tape. By scaling this triangle, we can see that each side would measure 100m, and the hypotenuse would measure approximately 141.42m (approximating the square root of 20,000). Thus, the maximum area would be a right-angled triangle, which is approximately 5,000 square meters.
So, depending on the scenario, you can enclose an area of either 5,625 or 5,000 square meters, Officer. I hope this helps you tape up the situation swiftly and securely!
To find the maximum area that can be enclosed, we need to determine the dimensions of the rectangular field.
Let's assume the length of the field is represented by 'L', and the width of the field is represented by 'W'.
Since there are three sides that need to be sealed, the total length of the yellow police tape used will be the perimeter of the field.
Perimeter = 2L + W
Given that the total length of the yellow police tape used is 300m, we can write the equation:
2L + W = 300
To find the maximum area, we need to express one variable in terms of the other. Let's solve the above equation for 'W' in terms of 'L':
W = 300 - 2L
Now, we can express the area of the rectangular field in terms of 'L':
Area = Length x Width
= L x (300 - 2L)
= 300L - 2L^2
To find the maximum area, we can take the derivative of the area function with respect to 'L' and set it equal to zero:
d(Area)/dL = 300 - 4L = 0
Solving for 'L':
300 = 4L
L = 75
Now, we can substitute the value of 'L' back into the equation for 'W':
W = 300 - 2L
W = 300 - 2(75)
W = 150
Therefore, the dimensions of the area that will enclose the maximum area are 75m x 150m, and the maximum area that can be enclosed is 75m x 150m = 11,250 square meters.
To find the maximum area that can be enclosed with 300 m of police tape, we need to determine the dimensions of the rectangular field.
Let's assume the length of the field is L and the width is W.
The police officer wants to enclose three sides of the field, which means two equal sides and one longer side. The total length of the two equal sides will be 2W, and the longer side will be L. So, the total length of the police tape required will be:
2W + L = 300 m (equation 1)
We need to find the maximum area, which is given by the formula A = L * W.
To solve this problem, we can use a system of equations. We have one equation (equation 1), and we need another equation relating the length and width.
Since the rectangular field is next to a building, we can assume that one side of the field is already connected to the building. This means one side's length is fixed, and we only need to find the other side's length.
Let's assume the length of the field connected to the building is B.
From the problem statement, we know that the total length of the field plus the tape is 300 m. So,
B + W + L + W = 300 m
2W + L + B = 300 m (equation 2)
Now we have two equations: equation 1 and equation 2.
We can solve these equations simultaneously to find the values of W and L:
2W + L = 300 m (equation 1)
2W + L + B = 300 m (equation 2)
To solve for W and L, we need another equation. Looking at the given information, we see that the field is rectangular, so opposite sides are equal in length. This means L = B.
Substituting L with B in equation 2:
2W + B + B = 300 m
2W + 2B = 300 m (equation 3)
Now we have two equations with two variables:
2W + L = 300 m (equation 1)
2W + 2B = 300 m (equation 3)
Solving these equations will give us the values of W and B, which we can use to find L. Once we have the dimensions, we can calculate the maximum area.
I can help you solve these equations. Just let me know the exact lengths or any other information you have, and I'll guide you through the calculations.