Answer: Let's start by finding the dimensions of the original rectangular room.
We know that the area of the room is 60m², so we can write:
length x width = 60
We want to find the length and width of the original room, so let's call them "L" and "W", respectively. Then we can write:
L x W = 60
Now let's consider the modified room, where the length is 25% less and the width is 4m more. If the room is a square, that means the length and width are equal. Let's call this side length "S". Then we can write:
S x S = (L - 0.25L) x (W + 4)
Simplifying, we get:
S² = 0.75LW + 4(L - 0.25L)
Simplifying further, we get:
S² = 0.75LW + 3L
Now we can use the fact that LW = 60 to eliminate one variable. Substituting, we get:
S² = 45 + 3L
We want to find the length of the longest stick that can be put on the floor of the original room. This is the diagonal of the rectangle, which we can find using the Pythagorean theorem:
diagonal² = L² + W²
Substituting LW = 60, we get:
diagonal² = L² + (60/L)²
To maximize the diagonal, we can take the derivative of the right-hand side with respect to L and set it equal to zero:
d/dL (L² + (60/L)²) = 2L - 7200/L³ = 0
Solving for L, we get:
L³ = 3600
L = 15
Therefore, the length of the longest stick that can be put on the floor of the room is the diagonal, which is:
diagonal = sqrt(15² + (60/15)²) = sqrt(225 + 16) = sqrt(241) ≈ 15.52 meters
Explanation: