can you please help me with this question, i'm completely lost.
talia's uncle owns a warehouse and he has given Talia an area in which to store the computer supplies for his company. her uncle gave Talia 40 m of rope and told her to section off a rectangular area in a corner of the warehouse. conduct an investigation to determine the greatest area that Talia can rope off.
please help me!! please!!!
A square would measure 20 meters on each side -- for an area of 400 square meters.
A rectangle that's 10 by 30 would give an area of 300 square meters.
If you assume that two sides of the area are bordered by walls and the other two sides by rope, the above would be correct.
However, for the square to use 40 m of rope, it would have 10 m on each of the four sides, giving an area of 100 sq. m.
For the rectangle, a rectangle could be 5x15 m (5+5+15+15), giving 75 sq m. of area.
I hope this helps a little more. Thanks for asking.
How about the triangle formed by the wall and the rope giving the hypotenuse? Ms Sue
40*40=1600
1600/4=400m2
The greatest area, 400 m2, is enclosed when
the length and width are each 20 m.
Of course, I'm here to help! To determine the greatest area that Talia can rope off in the warehouse, we need to find the dimensions of the rectangle that will maximize the area.
Let's start by understanding the scenario. Talia's uncle gave her 40 m of rope, which means she can use that rope to form the perimeter of the rectangle. Since the rope is used to form the perimeter of the rectangle, we can see that the length of the rope is equal to the sum of all sides of the rectangle.
So, we can start by understanding the formula for the perimeter of a rectangle:
Perimeter = 2 * (length + width)
In this case, since the rectangle is formed in a corner of the warehouse, we can consider it as an L-shape, with two sides adjacent to the walls of the warehouse. Let's call the length of the rectangle "L" and the width "W".
The perimeter equation now becomes:
40 = 2 * (L + W)
We can simplify this equation further:
20 = L + W
Now, to maximize the area, we need to find the dimensions of the rectangle that satisfy this equation while also maximizing the product of length and width (area = length * width).
There are various methods to solve this problem, but one approach is to plot the possible combinations of L and W that satisfy the equation and calculate the corresponding areas. Let's consider a few scenarios:
Scenario 1: L = 10 and W = 10
The perimeter equation is satisfied: 20 = 10 + 10
The area is: 10 * 10 = 100
Scenario 2: L = 5 and W = 15
The perimeter equation is satisfied: 20 = 5 + 15
The area is: 5 * 15 = 75
Scenario 3: L = 15 and W = 5
The perimeter equation is satisfied: 20 = 15 + 5
The area is: 15 * 5 = 75
By trying out a few different scenarios, we can observe that Scenario 1 gives us the largest area, with an area of 100 square meters.
Therefore, the greatest area that Talia can rope off in the warehouse is 100 square meters. She can achieve this by creating a rectangle with dimensions of 10 meters by 10 meters.